(A) The molar mass of $Na_2CO_3$ is $(2 \times 23) + 12 + (3 \times 16) = 106 \ g/mol$. The formula for molarity $(M)$ is $M = \frac{w \times 1000}{m.

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(A) The molar mass of $Na_2CO_3$ is $(2 \times 23) + 12 + (3 \times 16) = 106 \ g/mol$. The formula for molarity $(M)$ is $M = \frac{w \times 1000}{m.wt. \times V \text{ (in mL)}}$. Substituting the given values: $M = \frac{10.6 \times 1000}{106 \times 500} = \frac{10600}{53000} = 0.2 \ M$.

Class 12 Chemistry Solutions Method of expressing concentration of solution

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The molarity of a solution of $Na_2CO_3$ having $10.6 \ g$ in $500 \ mL$ of solution is ......... $M$.

A
$0.2$
B
$2$
C
$20$
D
$0.02$

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Method of expressing concentration of solution

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The molarity of a solution of $Na_2CO_3$ having $10.6 \ g$ in $500 \ mL$ of solution is ......... $M$.

Similar Questions

How many grams of $NaOH$ must be dissolved in $500 \, g$ of solution to obtain a $10 \% \, w/w$ concentration?

When $1.80 \ g$ of glucose is dissolved in $90 \ g$ of $H_2O$,the mole fraction of glucose is:

What volume of water is to be added to $50 \, mL$ of $4 \, N$ $NaOH$ solution to obtain a $1 \, N$ solution? $(... \, mL)$

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