(D) First,calculate the molarity of the stock solution:$M = \frac{\text{density} \times 10 \times \% (w/w)}{\text{molar mass}} = \frac{1.25 \times 10

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(D) First,calculate the molarity of the stock solution:$M = \frac{\text{density} \times 10 \times \% (w/w)}{\text{molar mass}} = \frac{1.25 \times 10 \times 29.2}{36.5} = 10 \ M$.Now,use the dilution formula $M_1V_1 = M_2V_2$:$10 \ M \times V_1 = 0.4 \ M \times 200 \ mL$.$V_1 = \frac{0.4 \times 200}{10} = 8 \ mL$.

Class 11 Chemistry Some Basic Concepts of Chemistry Chemical stoichiometry

$29.2 \% (w/w)$ $HCl$ stock solution has a density of $1.25 \ g \ mL^{-1}$. The molecular weight of $HCl$ is $36.5 \ g \ mol^{-1}$. The volume $(mL)$ of stock solution required to prepare a $200 \ mL$ solution of $0.4 \ M$ $HCl$ is :

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$5$
B
$6$
C
$7$
D
$8$

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Some Basic Concepts of Chemistry

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Chemical stoichiometry

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$29.2 \% (w/w)$ $HCl$ stock solution has a density of $1.25 \ g \ mL^{-1}$. The molecular weight of $HCl$ is $36.5 \ g \ mol^{-1}$. The volume $(mL)$ of stock solution required to prepare a $200 \ mL$ solution of $0.4 \ M$ $HCl$ is :

Similar Questions

How many moles of $H_2O$ (per mole of $Cl_2$ evolved) will be formed when $KMnO_4$ is added to $HCl$ solution?

The number of chlorine atoms in $20 \, mL$ of chlorine gas at $STP$ is ..... $\times 10^{21}$. (Round off to the Nearest Integer).[Assume chlorine is an ideal gas at $STP$,$R = 0.083 \, L \, bar \, mol^{-1} \, K^{-1}$,$N_{A} = 6.023 \times 10^{23}$]

Calculate the mass of sodium acetate $(CH_{3}COONa)$ required to make $500 \,mL$ of $0.375 \,M$ aqueous solution. Molar mass of sodium acetate is $82.0245 \,g \,mol^{-1}$.

If $5.85 \ g$ of sodium chloride is added to water to make a $500 \ mL$ solution,what is the molarity of the sodium chloride solution (in $M$)? (Molar mass $= 58.5 \ g/mol$)

If the percentage purity of an impure sample of a compound is given,then the weight of the substance required to prepare its $0.1 \, N$ solution will be:

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The number of chlorine atoms in $20 \, mL$ of chlorine gas at $STP$ is ..... $\times 10^{21}$. (Round off to the Nearest Integer).
[Assume chlorine is an ideal gas at $STP$,$R = 0.083 \, L \, bar \, mol^{-1} \, K^{-1}$,$N_{A} = 6.023 \times 10^{23}$]