319172
500 g of a solution of \({\mathrm{\mathrm{KMnO}_{4}}}\) labelled \({\mathrm{5 \% \mathrm{~W} / \mathrm{W}}}\) will contain ____ g of \({\mathrm{\mathrm{KMnO}_{4}}}\).
1 25
2 10
3 15
4 30
Explanation:
\({\text{5% }}\;{\text{(W/W)}}\) means 100 g solution contains 5 g \({\mathrm{\mathrm{KMnO}_{4}}}\). Hence, 500 g solution contains 25 g \({\text{KMn}}{{\text{O}}_{\text{4}}}\).
CHXII02:SOLUTIONS
319173
A solution is prepared by adding 2 g of " \({\text{X}}\) " to 1 mole of water. Mass percent of " \({\text{X}}\) " in the solution is
1 \(5 \%\)
2 \(10 \%\)
3 \(2 \%\)
4 \(20 \%\)
Explanation:
Given amount of solute \(({\rm{X}}) = 2\,{\rm{g}}\) Amount of solvent \(\left(\mathrm{H}_{2} \mathrm{O}\right)=1\) mole \(=18 \mathrm{~g}\) Total mass of solution \(=2+18=20 \mathrm{~g}\) Hence, mass percentage of \({\rm{X}} = \frac{2}{{20}} \times 100 = 10\,\% \) Thus, mass percent of " \({\text{X}}\) " in the solution is \(10 \%\).
JEE - 2023
CHXII02:SOLUTIONS
319174
A solution is obtained by mixing 300g of \(25\% \) solution and 400g of \(40\% \) solution by mass. The mass percentage of the resulting solution is:
1 \(66.66\% \)
2 \(3.36\% \)
3 \(33.6\% \)
4 \(22.4\% \)
Explanation:
Mass percentage of solute in the solution
CHXII02:SOLUTIONS
319175
Assertion : One molal aqueous solution of urea contains \(60 \mathrm{~g}\) of urea in \(1 \mathrm{~kg}\) of water. Reason : Solution containing one mole of solute in \(1000 \mathrm{~g}\) solvent is called one molal solution.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
Molecular weight of urea \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)\) is 60. \(\mathrm{g} / \mathrm{mole}\). Number of moles \(=\dfrac{\text { Weight }}{\text { Molecular weight }}=\dfrac{60}{60}=1\) \(\mathrm{m}=\dfrac{\text { No.of moles of solute }}{\text { Mass of solvent in } \mathrm{kg}}=\dfrac{1 \mathrm{~mol}}{1 \mathrm{~kg}}=1 \mathrm{~m}\) \(\therefore 1 \mathrm{~mol}\) (or) \(60 \mathrm{~g}\) of urea is present in \(1 \mathrm{~kg}\) of solvent. So option (1) is correct.
CHXII02:SOLUTIONS
319176
What is the mass ratio of ethylene glycol \({\text{(}}{{\text{C}}_2}{{\text{H}}_6}{{\text{O}}_2},\) molar mass \(\left.=62 \mathrm{~g} / \mathrm{mol}\right)\) required for making 500 g of 0.25 molal aqueous solution and 250 mL of 0.25 molal aqueous solution?
1 \(1: 2\)
2 \(1: 1\)
3 \(3: 1\)
4 \(2: 1\)
Explanation:
Assume: Mass of solvent \(\approx\) Mass of solution 0.25 molal \(=\dfrac{\text { No. of moles }}{500 \mathrm{~g}} \times 1000\) No. of moles \(=\dfrac{0.25}{2}\) \(\dfrac{\text { Mass }}{\text { Molar mass }}=\dfrac{0.25}{2}\) \(\Rightarrow(\text { Mass })_{1}=\dfrac{0.25}{2} \times 62=7.75 \mathrm{~g}\) \((\because\) density of water \(=1 \mathrm{~g} / \mathrm{cc}, 1 \mathrm{~g}=1 \mathrm{cc})\) 0.25 molal \(=\dfrac{\text { No. of } \text { moles }}{250(\mathrm{~mL})} \times 1000\) No. of moles \(=\dfrac{0.25}{4}\) \(\dfrac{\text { Mass }}{\text { Molar mass }}=\dfrac{0.25}{4}\) \(\Rightarrow(\text { Mass })_{2}=\dfrac{0.25}{4} \times 62=3.875 \mathrm{~g}\) Mass \(_{1}:\) Mass \(_{2}=7.75: 3.875=2: 1\)
319172
500 g of a solution of \({\mathrm{\mathrm{KMnO}_{4}}}\) labelled \({\mathrm{5 \% \mathrm{~W} / \mathrm{W}}}\) will contain ____ g of \({\mathrm{\mathrm{KMnO}_{4}}}\).
1 25
2 10
3 15
4 30
Explanation:
\({\text{5% }}\;{\text{(W/W)}}\) means 100 g solution contains 5 g \({\mathrm{\mathrm{KMnO}_{4}}}\). Hence, 500 g solution contains 25 g \({\text{KMn}}{{\text{O}}_{\text{4}}}\).
CHXII02:SOLUTIONS
319173
A solution is prepared by adding 2 g of " \({\text{X}}\) " to 1 mole of water. Mass percent of " \({\text{X}}\) " in the solution is
1 \(5 \%\)
2 \(10 \%\)
3 \(2 \%\)
4 \(20 \%\)
Explanation:
Given amount of solute \(({\rm{X}}) = 2\,{\rm{g}}\) Amount of solvent \(\left(\mathrm{H}_{2} \mathrm{O}\right)=1\) mole \(=18 \mathrm{~g}\) Total mass of solution \(=2+18=20 \mathrm{~g}\) Hence, mass percentage of \({\rm{X}} = \frac{2}{{20}} \times 100 = 10\,\% \) Thus, mass percent of " \({\text{X}}\) " in the solution is \(10 \%\).
JEE - 2023
CHXII02:SOLUTIONS
319174
A solution is obtained by mixing 300g of \(25\% \) solution and 400g of \(40\% \) solution by mass. The mass percentage of the resulting solution is:
1 \(66.66\% \)
2 \(3.36\% \)
3 \(33.6\% \)
4 \(22.4\% \)
Explanation:
Mass percentage of solute in the solution
CHXII02:SOLUTIONS
319175
Assertion : One molal aqueous solution of urea contains \(60 \mathrm{~g}\) of urea in \(1 \mathrm{~kg}\) of water. Reason : Solution containing one mole of solute in \(1000 \mathrm{~g}\) solvent is called one molal solution.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
Molecular weight of urea \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)\) is 60. \(\mathrm{g} / \mathrm{mole}\). Number of moles \(=\dfrac{\text { Weight }}{\text { Molecular weight }}=\dfrac{60}{60}=1\) \(\mathrm{m}=\dfrac{\text { No.of moles of solute }}{\text { Mass of solvent in } \mathrm{kg}}=\dfrac{1 \mathrm{~mol}}{1 \mathrm{~kg}}=1 \mathrm{~m}\) \(\therefore 1 \mathrm{~mol}\) (or) \(60 \mathrm{~g}\) of urea is present in \(1 \mathrm{~kg}\) of solvent. So option (1) is correct.
CHXII02:SOLUTIONS
319176
What is the mass ratio of ethylene glycol \({\text{(}}{{\text{C}}_2}{{\text{H}}_6}{{\text{O}}_2},\) molar mass \(\left.=62 \mathrm{~g} / \mathrm{mol}\right)\) required for making 500 g of 0.25 molal aqueous solution and 250 mL of 0.25 molal aqueous solution?
1 \(1: 2\)
2 \(1: 1\)
3 \(3: 1\)
4 \(2: 1\)
Explanation:
Assume: Mass of solvent \(\approx\) Mass of solution 0.25 molal \(=\dfrac{\text { No. of moles }}{500 \mathrm{~g}} \times 1000\) No. of moles \(=\dfrac{0.25}{2}\) \(\dfrac{\text { Mass }}{\text { Molar mass }}=\dfrac{0.25}{2}\) \(\Rightarrow(\text { Mass })_{1}=\dfrac{0.25}{2} \times 62=7.75 \mathrm{~g}\) \((\because\) density of water \(=1 \mathrm{~g} / \mathrm{cc}, 1 \mathrm{~g}=1 \mathrm{cc})\) 0.25 molal \(=\dfrac{\text { No. of } \text { moles }}{250(\mathrm{~mL})} \times 1000\) No. of moles \(=\dfrac{0.25}{4}\) \(\dfrac{\text { Mass }}{\text { Molar mass }}=\dfrac{0.25}{4}\) \(\Rightarrow(\text { Mass })_{2}=\dfrac{0.25}{4} \times 62=3.875 \mathrm{~g}\) Mass \(_{1}:\) Mass \(_{2}=7.75: 3.875=2: 1\)
319172
500 g of a solution of \({\mathrm{\mathrm{KMnO}_{4}}}\) labelled \({\mathrm{5 \% \mathrm{~W} / \mathrm{W}}}\) will contain ____ g of \({\mathrm{\mathrm{KMnO}_{4}}}\).
1 25
2 10
3 15
4 30
Explanation:
\({\text{5% }}\;{\text{(W/W)}}\) means 100 g solution contains 5 g \({\mathrm{\mathrm{KMnO}_{4}}}\). Hence, 500 g solution contains 25 g \({\text{KMn}}{{\text{O}}_{\text{4}}}\).
CHXII02:SOLUTIONS
319173
A solution is prepared by adding 2 g of " \({\text{X}}\) " to 1 mole of water. Mass percent of " \({\text{X}}\) " in the solution is
1 \(5 \%\)
2 \(10 \%\)
3 \(2 \%\)
4 \(20 \%\)
Explanation:
Given amount of solute \(({\rm{X}}) = 2\,{\rm{g}}\) Amount of solvent \(\left(\mathrm{H}_{2} \mathrm{O}\right)=1\) mole \(=18 \mathrm{~g}\) Total mass of solution \(=2+18=20 \mathrm{~g}\) Hence, mass percentage of \({\rm{X}} = \frac{2}{{20}} \times 100 = 10\,\% \) Thus, mass percent of " \({\text{X}}\) " in the solution is \(10 \%\).
JEE - 2023
CHXII02:SOLUTIONS
319174
A solution is obtained by mixing 300g of \(25\% \) solution and 400g of \(40\% \) solution by mass. The mass percentage of the resulting solution is:
1 \(66.66\% \)
2 \(3.36\% \)
3 \(33.6\% \)
4 \(22.4\% \)
Explanation:
Mass percentage of solute in the solution
CHXII02:SOLUTIONS
319175
Assertion : One molal aqueous solution of urea contains \(60 \mathrm{~g}\) of urea in \(1 \mathrm{~kg}\) of water. Reason : Solution containing one mole of solute in \(1000 \mathrm{~g}\) solvent is called one molal solution.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
Molecular weight of urea \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)\) is 60. \(\mathrm{g} / \mathrm{mole}\). Number of moles \(=\dfrac{\text { Weight }}{\text { Molecular weight }}=\dfrac{60}{60}=1\) \(\mathrm{m}=\dfrac{\text { No.of moles of solute }}{\text { Mass of solvent in } \mathrm{kg}}=\dfrac{1 \mathrm{~mol}}{1 \mathrm{~kg}}=1 \mathrm{~m}\) \(\therefore 1 \mathrm{~mol}\) (or) \(60 \mathrm{~g}\) of urea is present in \(1 \mathrm{~kg}\) of solvent. So option (1) is correct.
CHXII02:SOLUTIONS
319176
What is the mass ratio of ethylene glycol \({\text{(}}{{\text{C}}_2}{{\text{H}}_6}{{\text{O}}_2},\) molar mass \(\left.=62 \mathrm{~g} / \mathrm{mol}\right)\) required for making 500 g of 0.25 molal aqueous solution and 250 mL of 0.25 molal aqueous solution?
1 \(1: 2\)
2 \(1: 1\)
3 \(3: 1\)
4 \(2: 1\)
Explanation:
Assume: Mass of solvent \(\approx\) Mass of solution 0.25 molal \(=\dfrac{\text { No. of moles }}{500 \mathrm{~g}} \times 1000\) No. of moles \(=\dfrac{0.25}{2}\) \(\dfrac{\text { Mass }}{\text { Molar mass }}=\dfrac{0.25}{2}\) \(\Rightarrow(\text { Mass })_{1}=\dfrac{0.25}{2} \times 62=7.75 \mathrm{~g}\) \((\because\) density of water \(=1 \mathrm{~g} / \mathrm{cc}, 1 \mathrm{~g}=1 \mathrm{cc})\) 0.25 molal \(=\dfrac{\text { No. of } \text { moles }}{250(\mathrm{~mL})} \times 1000\) No. of moles \(=\dfrac{0.25}{4}\) \(\dfrac{\text { Mass }}{\text { Molar mass }}=\dfrac{0.25}{4}\) \(\Rightarrow(\text { Mass })_{2}=\dfrac{0.25}{4} \times 62=3.875 \mathrm{~g}\) Mass \(_{1}:\) Mass \(_{2}=7.75: 3.875=2: 1\)
319172
500 g of a solution of \({\mathrm{\mathrm{KMnO}_{4}}}\) labelled \({\mathrm{5 \% \mathrm{~W} / \mathrm{W}}}\) will contain ____ g of \({\mathrm{\mathrm{KMnO}_{4}}}\).
1 25
2 10
3 15
4 30
Explanation:
\({\text{5% }}\;{\text{(W/W)}}\) means 100 g solution contains 5 g \({\mathrm{\mathrm{KMnO}_{4}}}\). Hence, 500 g solution contains 25 g \({\text{KMn}}{{\text{O}}_{\text{4}}}\).
CHXII02:SOLUTIONS
319173
A solution is prepared by adding 2 g of " \({\text{X}}\) " to 1 mole of water. Mass percent of " \({\text{X}}\) " in the solution is
1 \(5 \%\)
2 \(10 \%\)
3 \(2 \%\)
4 \(20 \%\)
Explanation:
Given amount of solute \(({\rm{X}}) = 2\,{\rm{g}}\) Amount of solvent \(\left(\mathrm{H}_{2} \mathrm{O}\right)=1\) mole \(=18 \mathrm{~g}\) Total mass of solution \(=2+18=20 \mathrm{~g}\) Hence, mass percentage of \({\rm{X}} = \frac{2}{{20}} \times 100 = 10\,\% \) Thus, mass percent of " \({\text{X}}\) " in the solution is \(10 \%\).
JEE - 2023
CHXII02:SOLUTIONS
319174
A solution is obtained by mixing 300g of \(25\% \) solution and 400g of \(40\% \) solution by mass. The mass percentage of the resulting solution is:
1 \(66.66\% \)
2 \(3.36\% \)
3 \(33.6\% \)
4 \(22.4\% \)
Explanation:
Mass percentage of solute in the solution
CHXII02:SOLUTIONS
319175
Assertion : One molal aqueous solution of urea contains \(60 \mathrm{~g}\) of urea in \(1 \mathrm{~kg}\) of water. Reason : Solution containing one mole of solute in \(1000 \mathrm{~g}\) solvent is called one molal solution.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
Molecular weight of urea \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)\) is 60. \(\mathrm{g} / \mathrm{mole}\). Number of moles \(=\dfrac{\text { Weight }}{\text { Molecular weight }}=\dfrac{60}{60}=1\) \(\mathrm{m}=\dfrac{\text { No.of moles of solute }}{\text { Mass of solvent in } \mathrm{kg}}=\dfrac{1 \mathrm{~mol}}{1 \mathrm{~kg}}=1 \mathrm{~m}\) \(\therefore 1 \mathrm{~mol}\) (or) \(60 \mathrm{~g}\) of urea is present in \(1 \mathrm{~kg}\) of solvent. So option (1) is correct.
CHXII02:SOLUTIONS
319176
What is the mass ratio of ethylene glycol \({\text{(}}{{\text{C}}_2}{{\text{H}}_6}{{\text{O}}_2},\) molar mass \(\left.=62 \mathrm{~g} / \mathrm{mol}\right)\) required for making 500 g of 0.25 molal aqueous solution and 250 mL of 0.25 molal aqueous solution?
1 \(1: 2\)
2 \(1: 1\)
3 \(3: 1\)
4 \(2: 1\)
Explanation:
Assume: Mass of solvent \(\approx\) Mass of solution 0.25 molal \(=\dfrac{\text { No. of moles }}{500 \mathrm{~g}} \times 1000\) No. of moles \(=\dfrac{0.25}{2}\) \(\dfrac{\text { Mass }}{\text { Molar mass }}=\dfrac{0.25}{2}\) \(\Rightarrow(\text { Mass })_{1}=\dfrac{0.25}{2} \times 62=7.75 \mathrm{~g}\) \((\because\) density of water \(=1 \mathrm{~g} / \mathrm{cc}, 1 \mathrm{~g}=1 \mathrm{cc})\) 0.25 molal \(=\dfrac{\text { No. of } \text { moles }}{250(\mathrm{~mL})} \times 1000\) No. of moles \(=\dfrac{0.25}{4}\) \(\dfrac{\text { Mass }}{\text { Molar mass }}=\dfrac{0.25}{4}\) \(\Rightarrow(\text { Mass })_{2}=\dfrac{0.25}{4} \times 62=3.875 \mathrm{~g}\) Mass \(_{1}:\) Mass \(_{2}=7.75: 3.875=2: 1\)
319172
500 g of a solution of \({\mathrm{\mathrm{KMnO}_{4}}}\) labelled \({\mathrm{5 \% \mathrm{~W} / \mathrm{W}}}\) will contain ____ g of \({\mathrm{\mathrm{KMnO}_{4}}}\).
1 25
2 10
3 15
4 30
Explanation:
\({\text{5% }}\;{\text{(W/W)}}\) means 100 g solution contains 5 g \({\mathrm{\mathrm{KMnO}_{4}}}\). Hence, 500 g solution contains 25 g \({\text{KMn}}{{\text{O}}_{\text{4}}}\).
CHXII02:SOLUTIONS
319173
A solution is prepared by adding 2 g of " \({\text{X}}\) " to 1 mole of water. Mass percent of " \({\text{X}}\) " in the solution is
1 \(5 \%\)
2 \(10 \%\)
3 \(2 \%\)
4 \(20 \%\)
Explanation:
Given amount of solute \(({\rm{X}}) = 2\,{\rm{g}}\) Amount of solvent \(\left(\mathrm{H}_{2} \mathrm{O}\right)=1\) mole \(=18 \mathrm{~g}\) Total mass of solution \(=2+18=20 \mathrm{~g}\) Hence, mass percentage of \({\rm{X}} = \frac{2}{{20}} \times 100 = 10\,\% \) Thus, mass percent of " \({\text{X}}\) " in the solution is \(10 \%\).
JEE - 2023
CHXII02:SOLUTIONS
319174
A solution is obtained by mixing 300g of \(25\% \) solution and 400g of \(40\% \) solution by mass. The mass percentage of the resulting solution is:
1 \(66.66\% \)
2 \(3.36\% \)
3 \(33.6\% \)
4 \(22.4\% \)
Explanation:
Mass percentage of solute in the solution
CHXII02:SOLUTIONS
319175
Assertion : One molal aqueous solution of urea contains \(60 \mathrm{~g}\) of urea in \(1 \mathrm{~kg}\) of water. Reason : Solution containing one mole of solute in \(1000 \mathrm{~g}\) solvent is called one molal solution.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
Molecular weight of urea \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)\) is 60. \(\mathrm{g} / \mathrm{mole}\). Number of moles \(=\dfrac{\text { Weight }}{\text { Molecular weight }}=\dfrac{60}{60}=1\) \(\mathrm{m}=\dfrac{\text { No.of moles of solute }}{\text { Mass of solvent in } \mathrm{kg}}=\dfrac{1 \mathrm{~mol}}{1 \mathrm{~kg}}=1 \mathrm{~m}\) \(\therefore 1 \mathrm{~mol}\) (or) \(60 \mathrm{~g}\) of urea is present in \(1 \mathrm{~kg}\) of solvent. So option (1) is correct.
CHXII02:SOLUTIONS
319176
What is the mass ratio of ethylene glycol \({\text{(}}{{\text{C}}_2}{{\text{H}}_6}{{\text{O}}_2},\) molar mass \(\left.=62 \mathrm{~g} / \mathrm{mol}\right)\) required for making 500 g of 0.25 molal aqueous solution and 250 mL of 0.25 molal aqueous solution?
1 \(1: 2\)
2 \(1: 1\)
3 \(3: 1\)
4 \(2: 1\)
Explanation:
Assume: Mass of solvent \(\approx\) Mass of solution 0.25 molal \(=\dfrac{\text { No. of moles }}{500 \mathrm{~g}} \times 1000\) No. of moles \(=\dfrac{0.25}{2}\) \(\dfrac{\text { Mass }}{\text { Molar mass }}=\dfrac{0.25}{2}\) \(\Rightarrow(\text { Mass })_{1}=\dfrac{0.25}{2} \times 62=7.75 \mathrm{~g}\) \((\because\) density of water \(=1 \mathrm{~g} / \mathrm{cc}, 1 \mathrm{~g}=1 \mathrm{cc})\) 0.25 molal \(=\dfrac{\text { No. of } \text { moles }}{250(\mathrm{~mL})} \times 1000\) No. of moles \(=\dfrac{0.25}{4}\) \(\dfrac{\text { Mass }}{\text { Molar mass }}=\dfrac{0.25}{4}\) \(\Rightarrow(\text { Mass })_{2}=\dfrac{0.25}{4} \times 62=3.875 \mathrm{~g}\) Mass \(_{1}:\) Mass \(_{2}=7.75: 3.875=2: 1\)
