NEET Test Series from KOTA - 10 Papers In MS WORD
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1 RBTS PAPER
160621
How many grams of concentratd nitric acid solution should be used to prepare \(250 \mathrm{~mL}\) of \(2 \mathrm{M}\) \(\mathrm{HNO}_3\) ? The concentrated nitric acid is \(70 \% \mathrm{HNO}_3\) :
160622
What is the weight percentage of \(\mathrm{NaCl}\) solution in which \(\mathbf{2 0 ~ g ~ N a C l}\) is dissolved in \(\mathbf{6 0} \mathrm{g}\) of water:
1 \(10 \%\)
2 \(5 \%\)
3 \(25 \%\)
4 \(15 \%\)
Explanation:
\(\begin{aligned} \% w / w & =\frac{\text { wt. of solute }}{\text { wt. of solution }} \times 100=\frac{20}{20+60} \times 100 \\ & =25 \% \end{aligned}\)
NCERT 19
1 RBTS PAPER
160623
What is weight/volume percentage of a solution in which \(7.5 \mathrm{~g}\) of \(\mathrm{KCl}\) is dissolved in \(100 \mathrm{~mL}\) of the solution:
160624
If a water sample contains \(6.02 \times 10^{23}\) electrons, the volume of sample will be :(Density of water 1.0 \(\mathrm{gm} / \mathrm{ml})\) :
1 \(1.8 \mathrm{ml}\)
2 \(0.18 \mathrm{ml}\)
3 \(0.018 \mathrm{ml}\)
4 \(0.0018 \mathrm{ml}\)
Explanation:
No. of moles of electron \(=\frac{6.02 \times 10^{23}}{6.02 \times 10^{2.3}}=1\) mole 10 mole electron \(=1\) mole \(\mathrm{H}_2 \mathrm{O}=18 \mathrm{gm}\)
160621
How many grams of concentratd nitric acid solution should be used to prepare \(250 \mathrm{~mL}\) of \(2 \mathrm{M}\) \(\mathrm{HNO}_3\) ? The concentrated nitric acid is \(70 \% \mathrm{HNO}_3\) :
160622
What is the weight percentage of \(\mathrm{NaCl}\) solution in which \(\mathbf{2 0 ~ g ~ N a C l}\) is dissolved in \(\mathbf{6 0} \mathrm{g}\) of water:
1 \(10 \%\)
2 \(5 \%\)
3 \(25 \%\)
4 \(15 \%\)
Explanation:
\(\begin{aligned} \% w / w & =\frac{\text { wt. of solute }}{\text { wt. of solution }} \times 100=\frac{20}{20+60} \times 100 \\ & =25 \% \end{aligned}\)
NCERT 19
1 RBTS PAPER
160623
What is weight/volume percentage of a solution in which \(7.5 \mathrm{~g}\) of \(\mathrm{KCl}\) is dissolved in \(100 \mathrm{~mL}\) of the solution:
160624
If a water sample contains \(6.02 \times 10^{23}\) electrons, the volume of sample will be :(Density of water 1.0 \(\mathrm{gm} / \mathrm{ml})\) :
1 \(1.8 \mathrm{ml}\)
2 \(0.18 \mathrm{ml}\)
3 \(0.018 \mathrm{ml}\)
4 \(0.0018 \mathrm{ml}\)
Explanation:
No. of moles of electron \(=\frac{6.02 \times 10^{23}}{6.02 \times 10^{2.3}}=1\) mole 10 mole electron \(=1\) mole \(\mathrm{H}_2 \mathrm{O}=18 \mathrm{gm}\)
160621
How many grams of concentratd nitric acid solution should be used to prepare \(250 \mathrm{~mL}\) of \(2 \mathrm{M}\) \(\mathrm{HNO}_3\) ? The concentrated nitric acid is \(70 \% \mathrm{HNO}_3\) :
160622
What is the weight percentage of \(\mathrm{NaCl}\) solution in which \(\mathbf{2 0 ~ g ~ N a C l}\) is dissolved in \(\mathbf{6 0} \mathrm{g}\) of water:
1 \(10 \%\)
2 \(5 \%\)
3 \(25 \%\)
4 \(15 \%\)
Explanation:
\(\begin{aligned} \% w / w & =\frac{\text { wt. of solute }}{\text { wt. of solution }} \times 100=\frac{20}{20+60} \times 100 \\ & =25 \% \end{aligned}\)
NCERT 19
1 RBTS PAPER
160623
What is weight/volume percentage of a solution in which \(7.5 \mathrm{~g}\) of \(\mathrm{KCl}\) is dissolved in \(100 \mathrm{~mL}\) of the solution:
160624
If a water sample contains \(6.02 \times 10^{23}\) electrons, the volume of sample will be :(Density of water 1.0 \(\mathrm{gm} / \mathrm{ml})\) :
1 \(1.8 \mathrm{ml}\)
2 \(0.18 \mathrm{ml}\)
3 \(0.018 \mathrm{ml}\)
4 \(0.0018 \mathrm{ml}\)
Explanation:
No. of moles of electron \(=\frac{6.02 \times 10^{23}}{6.02 \times 10^{2.3}}=1\) mole 10 mole electron \(=1\) mole \(\mathrm{H}_2 \mathrm{O}=18 \mathrm{gm}\)
160621
How many grams of concentratd nitric acid solution should be used to prepare \(250 \mathrm{~mL}\) of \(2 \mathrm{M}\) \(\mathrm{HNO}_3\) ? The concentrated nitric acid is \(70 \% \mathrm{HNO}_3\) :
160622
What is the weight percentage of \(\mathrm{NaCl}\) solution in which \(\mathbf{2 0 ~ g ~ N a C l}\) is dissolved in \(\mathbf{6 0} \mathrm{g}\) of water:
1 \(10 \%\)
2 \(5 \%\)
3 \(25 \%\)
4 \(15 \%\)
Explanation:
\(\begin{aligned} \% w / w & =\frac{\text { wt. of solute }}{\text { wt. of solution }} \times 100=\frac{20}{20+60} \times 100 \\ & =25 \% \end{aligned}\)
NCERT 19
1 RBTS PAPER
160623
What is weight/volume percentage of a solution in which \(7.5 \mathrm{~g}\) of \(\mathrm{KCl}\) is dissolved in \(100 \mathrm{~mL}\) of the solution:
160624
If a water sample contains \(6.02 \times 10^{23}\) electrons, the volume of sample will be :(Density of water 1.0 \(\mathrm{gm} / \mathrm{ml})\) :
1 \(1.8 \mathrm{ml}\)
2 \(0.18 \mathrm{ml}\)
3 \(0.018 \mathrm{ml}\)
4 \(0.0018 \mathrm{ml}\)
Explanation:
No. of moles of electron \(=\frac{6.02 \times 10^{23}}{6.02 \times 10^{2.3}}=1\) mole 10 mole electron \(=1\) mole \(\mathrm{H}_2 \mathrm{O}=18 \mathrm{gm}\)
