277541
Sodium hydride when dissolved in water produces :
1 Acidic solution
2 Basic solution
3 Neutral solution
4 Cannot say
Explanation:
The hydrolysis of sodium hydride occurs in the reaction where it gets broken into to $\mathrm{Na}^{+}$cation and $\mathrm{H}^{-}$anion. The $\mathrm{Na}^{+}$forms sodium hydroxide and $\mathrm{H}^{-}$ forms hydrogen gas. The formation of sodium hydroxide reduces the risk of fires. Sodium Hydroxide is corrosive in nature. The reaction takes place as follows: $\mathrm{NaH}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{NaOH}+\mathrm{H}_{2}(\mathrm{~g})$ The hydrolysis of $\mathrm{NaH}$ is violent. The reaction is an exothermic reaction in which the energy is released.
MPPET-2013
SOLUTIONS
277542
Which one of the following is an isotonic pair of solution?
1 $0.15 \mathrm{M} \mathrm{NaCl}$ and $0.1 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}$
2 $0.2 \mathrm{M}$ Urea and $0.1 \mathrm{M}$ Sugar
3 $0.1 \mathrm{M} \mathrm{BaCl}_{2}$ and $0.2 \mathrm{M}$ Urea
4 $0.4 \mathrm{M} \mathrm{MgSO}_{4}$ and $0.1 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}$
Explanation:
Two solution having the same osmotic pressure across a semi permeable membrane is referred to as isotonic solution. For $0.15 \mathrm{M} \mathrm{NaCl}$ and $0.1 \mathrm{M} \mathrm{Na} \mathrm{Na}_2 \mathrm{SO}_4, \mathrm{NaCl}$ is an electrolyte which dissociates to give 2 ions, thus concentration of ions in the solution $0.30 \mathrm{M}$. Similarly $\mathrm{Na}_{2} \mathrm{SO}_{4}$ (3 ions) the concentration of ions in the solution $=0.30 \mathrm{M}$ Hence, both are isotonic.
AP-EAMCET (Engg.) 2013
SOLUTIONS
277543
On adding $\mathrm{AgNO}_{3}$ solution into $\mathrm{KI}$ solution, a negatively charged colloidal sol is obtained when they are in-
1 $100 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{AgNO}_{3}+100 \mathrm{~mL}$ of $0.1 \mathrm{M}$ $\mathrm{KI}$
2 $100 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{AgNO}_{3}+100 \mathrm{~mL}$ of $0.2 \mathrm{M}$ $\mathrm{KI}$
3 $100 \mathrm{~mL}$ of $0.2 \mathrm{M} \mathrm{AgNO}_{3}+100 \mathrm{~mL}$ of $0.1 \mathrm{M}$ $\mathrm{KI}$
4 $100 \mathrm{~mL}$ of $0.15 \mathrm{M} \mathrm{AgNO}+100 \mathrm{~mL}$ of $0.15 \mathrm{M}$ $\mathrm{KI}$
Explanation:
Negatively charged colloidal solution is formed when $\mathrm{AgNO}_{3}$ is completely precipitated as $\mathrm{AgI}$ and extra $\mathrm{KI}$ is absorbed on $\mathrm{AgI}$. $\mathrm{Ag}^{+}+\mathrm{I}^{-} \rightarrow \mathrm{AgI}$ Thus, $[\mathrm{Ag}+]<\left[\mathrm{I}^{-}\right]$
BCECE-2013
SOLUTIONS
277546
A $6 \%$ solution of urea is isotonic with
1 $1 \mathrm{M}$ solution of glucose
2 $0.05 \mathrm{M}$ solution of glucose
3 $6 \%$ solution of glucose
4 $25 \%$ solution of glucose
Explanation:
Molarity of urea $=\frac{\frac{6}{60}}{\frac{100}{1000}}=1 \mathrm{M}$ Hence, $1 \mathrm{M}$ solution of glucose is isotonic with $6 \%$ urea solution.
277541
Sodium hydride when dissolved in water produces :
1 Acidic solution
2 Basic solution
3 Neutral solution
4 Cannot say
Explanation:
The hydrolysis of sodium hydride occurs in the reaction where it gets broken into to $\mathrm{Na}^{+}$cation and $\mathrm{H}^{-}$anion. The $\mathrm{Na}^{+}$forms sodium hydroxide and $\mathrm{H}^{-}$ forms hydrogen gas. The formation of sodium hydroxide reduces the risk of fires. Sodium Hydroxide is corrosive in nature. The reaction takes place as follows: $\mathrm{NaH}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{NaOH}+\mathrm{H}_{2}(\mathrm{~g})$ The hydrolysis of $\mathrm{NaH}$ is violent. The reaction is an exothermic reaction in which the energy is released.
MPPET-2013
SOLUTIONS
277542
Which one of the following is an isotonic pair of solution?
1 $0.15 \mathrm{M} \mathrm{NaCl}$ and $0.1 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}$
2 $0.2 \mathrm{M}$ Urea and $0.1 \mathrm{M}$ Sugar
3 $0.1 \mathrm{M} \mathrm{BaCl}_{2}$ and $0.2 \mathrm{M}$ Urea
4 $0.4 \mathrm{M} \mathrm{MgSO}_{4}$ and $0.1 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}$
Explanation:
Two solution having the same osmotic pressure across a semi permeable membrane is referred to as isotonic solution. For $0.15 \mathrm{M} \mathrm{NaCl}$ and $0.1 \mathrm{M} \mathrm{Na} \mathrm{Na}_2 \mathrm{SO}_4, \mathrm{NaCl}$ is an electrolyte which dissociates to give 2 ions, thus concentration of ions in the solution $0.30 \mathrm{M}$. Similarly $\mathrm{Na}_{2} \mathrm{SO}_{4}$ (3 ions) the concentration of ions in the solution $=0.30 \mathrm{M}$ Hence, both are isotonic.
AP-EAMCET (Engg.) 2013
SOLUTIONS
277543
On adding $\mathrm{AgNO}_{3}$ solution into $\mathrm{KI}$ solution, a negatively charged colloidal sol is obtained when they are in-
1 $100 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{AgNO}_{3}+100 \mathrm{~mL}$ of $0.1 \mathrm{M}$ $\mathrm{KI}$
2 $100 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{AgNO}_{3}+100 \mathrm{~mL}$ of $0.2 \mathrm{M}$ $\mathrm{KI}$
3 $100 \mathrm{~mL}$ of $0.2 \mathrm{M} \mathrm{AgNO}_{3}+100 \mathrm{~mL}$ of $0.1 \mathrm{M}$ $\mathrm{KI}$
4 $100 \mathrm{~mL}$ of $0.15 \mathrm{M} \mathrm{AgNO}+100 \mathrm{~mL}$ of $0.15 \mathrm{M}$ $\mathrm{KI}$
Explanation:
Negatively charged colloidal solution is formed when $\mathrm{AgNO}_{3}$ is completely precipitated as $\mathrm{AgI}$ and extra $\mathrm{KI}$ is absorbed on $\mathrm{AgI}$. $\mathrm{Ag}^{+}+\mathrm{I}^{-} \rightarrow \mathrm{AgI}$ Thus, $[\mathrm{Ag}+]<\left[\mathrm{I}^{-}\right]$
BCECE-2013
SOLUTIONS
277546
A $6 \%$ solution of urea is isotonic with
1 $1 \mathrm{M}$ solution of glucose
2 $0.05 \mathrm{M}$ solution of glucose
3 $6 \%$ solution of glucose
4 $25 \%$ solution of glucose
Explanation:
Molarity of urea $=\frac{\frac{6}{60}}{\frac{100}{1000}}=1 \mathrm{M}$ Hence, $1 \mathrm{M}$ solution of glucose is isotonic with $6 \%$ urea solution.
277541
Sodium hydride when dissolved in water produces :
1 Acidic solution
2 Basic solution
3 Neutral solution
4 Cannot say
Explanation:
The hydrolysis of sodium hydride occurs in the reaction where it gets broken into to $\mathrm{Na}^{+}$cation and $\mathrm{H}^{-}$anion. The $\mathrm{Na}^{+}$forms sodium hydroxide and $\mathrm{H}^{-}$ forms hydrogen gas. The formation of sodium hydroxide reduces the risk of fires. Sodium Hydroxide is corrosive in nature. The reaction takes place as follows: $\mathrm{NaH}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{NaOH}+\mathrm{H}_{2}(\mathrm{~g})$ The hydrolysis of $\mathrm{NaH}$ is violent. The reaction is an exothermic reaction in which the energy is released.
MPPET-2013
SOLUTIONS
277542
Which one of the following is an isotonic pair of solution?
1 $0.15 \mathrm{M} \mathrm{NaCl}$ and $0.1 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}$
2 $0.2 \mathrm{M}$ Urea and $0.1 \mathrm{M}$ Sugar
3 $0.1 \mathrm{M} \mathrm{BaCl}_{2}$ and $0.2 \mathrm{M}$ Urea
4 $0.4 \mathrm{M} \mathrm{MgSO}_{4}$ and $0.1 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}$
Explanation:
Two solution having the same osmotic pressure across a semi permeable membrane is referred to as isotonic solution. For $0.15 \mathrm{M} \mathrm{NaCl}$ and $0.1 \mathrm{M} \mathrm{Na} \mathrm{Na}_2 \mathrm{SO}_4, \mathrm{NaCl}$ is an electrolyte which dissociates to give 2 ions, thus concentration of ions in the solution $0.30 \mathrm{M}$. Similarly $\mathrm{Na}_{2} \mathrm{SO}_{4}$ (3 ions) the concentration of ions in the solution $=0.30 \mathrm{M}$ Hence, both are isotonic.
AP-EAMCET (Engg.) 2013
SOLUTIONS
277543
On adding $\mathrm{AgNO}_{3}$ solution into $\mathrm{KI}$ solution, a negatively charged colloidal sol is obtained when they are in-
1 $100 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{AgNO}_{3}+100 \mathrm{~mL}$ of $0.1 \mathrm{M}$ $\mathrm{KI}$
2 $100 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{AgNO}_{3}+100 \mathrm{~mL}$ of $0.2 \mathrm{M}$ $\mathrm{KI}$
3 $100 \mathrm{~mL}$ of $0.2 \mathrm{M} \mathrm{AgNO}_{3}+100 \mathrm{~mL}$ of $0.1 \mathrm{M}$ $\mathrm{KI}$
4 $100 \mathrm{~mL}$ of $0.15 \mathrm{M} \mathrm{AgNO}+100 \mathrm{~mL}$ of $0.15 \mathrm{M}$ $\mathrm{KI}$
Explanation:
Negatively charged colloidal solution is formed when $\mathrm{AgNO}_{3}$ is completely precipitated as $\mathrm{AgI}$ and extra $\mathrm{KI}$ is absorbed on $\mathrm{AgI}$. $\mathrm{Ag}^{+}+\mathrm{I}^{-} \rightarrow \mathrm{AgI}$ Thus, $[\mathrm{Ag}+]<\left[\mathrm{I}^{-}\right]$
BCECE-2013
SOLUTIONS
277546
A $6 \%$ solution of urea is isotonic with
1 $1 \mathrm{M}$ solution of glucose
2 $0.05 \mathrm{M}$ solution of glucose
3 $6 \%$ solution of glucose
4 $25 \%$ solution of glucose
Explanation:
Molarity of urea $=\frac{\frac{6}{60}}{\frac{100}{1000}}=1 \mathrm{M}$ Hence, $1 \mathrm{M}$ solution of glucose is isotonic with $6 \%$ urea solution.
277541
Sodium hydride when dissolved in water produces :
1 Acidic solution
2 Basic solution
3 Neutral solution
4 Cannot say
Explanation:
The hydrolysis of sodium hydride occurs in the reaction where it gets broken into to $\mathrm{Na}^{+}$cation and $\mathrm{H}^{-}$anion. The $\mathrm{Na}^{+}$forms sodium hydroxide and $\mathrm{H}^{-}$ forms hydrogen gas. The formation of sodium hydroxide reduces the risk of fires. Sodium Hydroxide is corrosive in nature. The reaction takes place as follows: $\mathrm{NaH}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{NaOH}+\mathrm{H}_{2}(\mathrm{~g})$ The hydrolysis of $\mathrm{NaH}$ is violent. The reaction is an exothermic reaction in which the energy is released.
MPPET-2013
SOLUTIONS
277542
Which one of the following is an isotonic pair of solution?
1 $0.15 \mathrm{M} \mathrm{NaCl}$ and $0.1 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}$
2 $0.2 \mathrm{M}$ Urea and $0.1 \mathrm{M}$ Sugar
3 $0.1 \mathrm{M} \mathrm{BaCl}_{2}$ and $0.2 \mathrm{M}$ Urea
4 $0.4 \mathrm{M} \mathrm{MgSO}_{4}$ and $0.1 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}$
Explanation:
Two solution having the same osmotic pressure across a semi permeable membrane is referred to as isotonic solution. For $0.15 \mathrm{M} \mathrm{NaCl}$ and $0.1 \mathrm{M} \mathrm{Na} \mathrm{Na}_2 \mathrm{SO}_4, \mathrm{NaCl}$ is an electrolyte which dissociates to give 2 ions, thus concentration of ions in the solution $0.30 \mathrm{M}$. Similarly $\mathrm{Na}_{2} \mathrm{SO}_{4}$ (3 ions) the concentration of ions in the solution $=0.30 \mathrm{M}$ Hence, both are isotonic.
AP-EAMCET (Engg.) 2013
SOLUTIONS
277543
On adding $\mathrm{AgNO}_{3}$ solution into $\mathrm{KI}$ solution, a negatively charged colloidal sol is obtained when they are in-
1 $100 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{AgNO}_{3}+100 \mathrm{~mL}$ of $0.1 \mathrm{M}$ $\mathrm{KI}$
2 $100 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{AgNO}_{3}+100 \mathrm{~mL}$ of $0.2 \mathrm{M}$ $\mathrm{KI}$
3 $100 \mathrm{~mL}$ of $0.2 \mathrm{M} \mathrm{AgNO}_{3}+100 \mathrm{~mL}$ of $0.1 \mathrm{M}$ $\mathrm{KI}$
4 $100 \mathrm{~mL}$ of $0.15 \mathrm{M} \mathrm{AgNO}+100 \mathrm{~mL}$ of $0.15 \mathrm{M}$ $\mathrm{KI}$
Explanation:
Negatively charged colloidal solution is formed when $\mathrm{AgNO}_{3}$ is completely precipitated as $\mathrm{AgI}$ and extra $\mathrm{KI}$ is absorbed on $\mathrm{AgI}$. $\mathrm{Ag}^{+}+\mathrm{I}^{-} \rightarrow \mathrm{AgI}$ Thus, $[\mathrm{Ag}+]<\left[\mathrm{I}^{-}\right]$
BCECE-2013
SOLUTIONS
277546
A $6 \%$ solution of urea is isotonic with
1 $1 \mathrm{M}$ solution of glucose
2 $0.05 \mathrm{M}$ solution of glucose
3 $6 \%$ solution of glucose
4 $25 \%$ solution of glucose
Explanation:
Molarity of urea $=\frac{\frac{6}{60}}{\frac{100}{1000}}=1 \mathrm{M}$ Hence, $1 \mathrm{M}$ solution of glucose is isotonic with $6 \%$ urea solution.
