(C) The molar mass of $NaOH$ is $23 + 16 + 1 = 40 \ g/mol$.Number of moles of $NaOH = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{4 \ g}{40 \

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(C) The molar mass of $NaOH$ is $23 + 16 + 1 = 40 \ g/mol$.Number of moles of $NaOH = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{4 \ g}{40 \ g/mol} = 0.1 \ mol$.The volume of the solution is $250 \ mL = 0.25 \ L$.Molarity $(M) = \frac{\text{Number of moles of solute}}{\text{Volume of solution in } L} = \frac{0.1 \ mol}{0.25 \ L} = 0.4 \ M$.

Class 12 Chemistry Solutions Method of expressing concentration of solution

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$4 \ g$ of $NaOH$ is dissolved in $250 \ mL$ of water to prepare a solution. Find the molarity of the solution. (in $M$)

A
$0.1$
B
$0.2$
C
$0.4$
D
$0.8$

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Method of expressing concentration of solution

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$4 \ g$ of $NaOH$ is dissolved in $250 \ mL$ of water to prepare a solution. Find the molarity of the solution. (in $M$)

Similar Questions

The percentage (by weight) of sodium hydroxide in a $1.25$ molal $NaOH$ solution is (in $\%$)

$A$ solution of a solute in benzene has a molality of $X \ m$. If the mole fraction of the solute is $0.2$,what is the value of $X$?

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