(B) Mass of $NaOH$ $(m) = 3.2 \ g$ Volume of solution $(V) = 250 \ cm^{3} = 0.25 \ L$ Molar mass of $NaOH$ $(M) = 40 \ g \ mol^{-1}$ Number of moles $

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(B) Mass of $NaOH$ $(m) = 3.2 \ g$ Volume of solution $(V) = 250 \ cm^{3} = 0.25 \ L$ Molar mass of $NaOH$ $(M) = 40 \ g \ mol^{-1}$ Number of moles $(n) = \frac{m}{M} = \frac{3.2 \ g}{40 \ g \ mol^{-1}} = 0.08 \ mol$ Molarity $(M) = \frac{n}{V(L)} = \frac{0.08 \ mol}{0.25 \ L} = 0.32 \ mol \ dm^{-3}$

Class 12 Chemistry Solutions Method of expressing concentration of solution

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What is the molarity of a solution containing $3.2 \ g$ of $NaOH$ (molar mass $40 \ g \ mol^{-1}$) in $250 \ cm^{3}$ of water?

MHT CET 2020 All MHT CET

A
$0.512 \ mol \ dm^{-3}$
B
$0.32 \ mol \ dm^{-3}$
C
$0.032 \ mol \ dm^{-3}$
D
$0.02 \ mol \ dm^{-3}$

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What is the molarity of a solution containing $3.2 \ g$ of $NaOH$ (molar mass $40 \ g \ mol^{-1}$) in $250 \ cm^{3}$ of water?

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$1000 \ g$ aqueous solution of $CaCO_3$ contains $10 \ g$ of carbonate. Concentration of the solution is .......... $ppm$.

The correct relationship between molality $(m)$ and molarity $(M)$ of a solution is ($d = $ density in $g/mL$ and $M_{solute} = $ molar mass of solute in $g/mol$)

The number of moles of a solute in its solution is $20$ and the total number of moles is $80$. The mole fraction of the solute is:

The density of $3 \ M$ solution of $NaCl$ is $1.25 \ g \ mL^{-1}$. Calculate the molality of the solution.

$A$ solution is prepared by adding $124 \ g$ of ethylene glycol (molar mass $= 62 \ g \ mol^{-1}$) to $x \ g$ of water to get a $10 \ m$ solution. What is the value of $x$ (in $g$)?

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