(A) $I$. Moles of $NaOH = \text{Molarity} \times \text{Volume (L)} = 0.2 \times 0.5 = 0.1 \ mol$.$II$. For $H_2SO_4$,$n$-factor $= 2$. Molarity $= \fr

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(A) $I$. Moles of $NaOH = \text{Molarity} \times \text{Volume (L)} = 0.2 \times 0.5 = 0.1 \ mol$.$II$. For $H_2SO_4$,$n$-factor $= 2$. Molarity $= \frac{\text{Normality}}{n-\text{factor}} = \frac{0.1}{2} = 0.05 \ M$.Moles of $H_2SO_4 = 0.05 \times 0.2 = 0.01 \ mol$.$III$. Molar mass of urea $(NH_2CONH_2) = 60 \ g/mol$.Moles of urea $= \frac{\text{Given mass}}{\text{Molar mass}} = \frac{6}{60} = 0.1 \ mol$.Thus,the number of moles are $0.1, 0.01, 0.1$. Hence,option $(A)$ is correct.

Class 11 Chemistry Some Basic Concepts of Chemistry Mix Examples of Some Basic Concept of Chemistry

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The number of moles of solute present in the solutions of $I$,$II$ and $III$ is respectively:
$I$. $500 \ mL$ of $0.2 \ M \ NaOH$
$II$. $200 \ mL$ of $0.1 \ N \ H_2SO_4$
$III$. $6 \ g$ of urea in $1 \ kg$ of water

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A
$0.1, 0.01, 0.1$
B
$0.1, 0.02, 0.1$
C
$0.2, 0.01, 0.1$
D
$0.1, 0.01, 0.2$

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The number of moles of solute present in the solutions of $I$,$II$ and $III$ is respectively:$I$. $500 \ mL$ of $0.2 \ M \ NaOH$$II$. $200 \ mL$ of $0.1 \ N \ H_2SO_4$$III$. $6 \ g$ of urea in $1 \ kg$ of water

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You are given $500 \ mL$ of $2 \ N \ HCl$ and $500 \ mL$ of $5 \ N \ HCl$. What will be the maximum volume of $3 \ M \ HCl$ that you can make from these two solutions? .............. $mL$

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The number of moles of solute present in the solutions of $I$,$II$ and $III$ is respectively:
$I$. $500 \ mL$ of $0.2 \ M \ NaOH$
$II$. $200 \ mL$ of $0.1 \ N \ H_2SO_4$
$III$. $6 \ g$ of urea in $1 \ kg$ of water