(D) The molar mass of $Na_2CO_3$ is $106 \, g/mol$. The molarity of the initial solution is calculated as: $M_1 = \frac{\text{mass}}{\text{molar mass}

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(D) The molar mass of $Na_2CO_3$ is $106 \, g/mol$. The molarity of the initial solution is calculated as: $M_1 = \frac{\text{mass}}{\text{molar mass} \times \text{volume in L}} = \frac{5.3}{106 \times 0.250} = 0.2 \, M$. Using the dilution formula $M_1 V_1 = M_2 V_2$: $0.2 \, M \times 10 \, mL = M_2 \times 100 \, mL$. $M_2 = \frac{0.2 \times 10}{100} = 0.02 \, M$.

Class 11 Chemistry Some Basic Concepts of Chemistry Chemical stoichiometry

Easy

$A$ $250 \, mL$ solution of sodium carbonate contains $5.3 \, g$ of $Na_2CO_3$. If $10 \, mL$ of this solution is taken and diluted with distilled water to $0.1 \, L$,what will be the concentration of the resulting solution?

A
$0.01 \, M$
B
$0.02 \, m$
C
$0.002 \, M$
D
$0.02 \, M$

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Some Basic Concepts of Chemistry

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Chemical stoichiometry

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$A$ $250 \, mL$ solution of sodium carbonate contains $5.3 \, g$ of $Na_2CO_3$. If $10 \, mL$ of this solution is taken and diluted with distilled water to $0.1 \, L$,what will be the concentration of the resulting solution?

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The charge carried by $1 \text{ millimole}$ of $M^{n+}$ ions is $193 \text{ C}$. The value of $n$ is

The molarity of $HNO_{3}$ in a sample which has density $1.4 \; g/mL$ and mass percentage of $63\%$ is . . . . . . $M$.(Molecular Weight of $HNO_{3} = 63$)

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The molarity of $HNO_{3}$ in a sample which has density $1.4 \; g/mL$ and mass percentage of $63\%$ is . . . . . . $M$.
(Molecular Weight of $HNO_{3} = 63$)