(C) The molarity $(M)$ of a solution is defined as the number of moles of solute $(n)$ per liter of solution ($V$ in $L$): $M = \frac{n}{V(L)}$ Given:

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(C) The molarity $(M)$ of a solution is defined as the number of moles of solute $(n)$ per liter of solution ($V$ in $L$): $M = \frac{n}{V(L)}$ Given: $M = 0.5 \ M$ $V = 2 \ L$ Substituting the values into the formula: $0.5 = \frac{n}{2}$ $n = 0.5 \times 2 = 1 \ mol$ Therefore,the number of moles is $1$.

Class 11 Chemistry Some Basic Concepts of Chemistry The mole concept

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The number of moles present in $2 \ L$ of $0.5 \ M$ $NaOH$ is

A
$0.5$
B
$0.1$
C
$1$
D
$2$

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Class 11

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Chapter

Some Basic Concepts of Chemistry

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The mole concept

Explain the terms: Avogadro constant,molar volume of $1$ mole of gas,and $STP$.

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$0.1$ mole of compound $'S'$ will weigh $.......... \ g$.
(Given molar mass in $g \ mol^{-1} \ C: 12, H: 1, O: 16$)

DifficultJEE MAIN 2025

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The number of water molecules present in a drop of water weighing $0.018 \ g$ is

DifficultKCET 2013

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How many moles are present in $0.224 \ L$ of an ideal $H_2$ gas at $STP$ (in $mol$)? (Assume molar volume $= 22.4 \ L \ mol^{-1}$)

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The number of moles present in $2 \ L$ of $0.5 \ M$ $NaOH$ is

Similar Questions

Explain the terms: Avogadro constant,molar volume of $1$ mole of gas,and $STP$.

The weight of $112 \ cm^3$ of $CH_4$ gas at $STP$ is ...... $g$.

$0.1$ mole of compound $'S'$ will weigh $.......... \ g$.(Given molar mass in $g \ mol^{-1} \ C: 12, H: 1, O: 16$)

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$0.1$ mole of compound $'S'$ will weigh $.......... \ g$.
(Given molar mass in $g \ mol^{-1} \ C: 12, H: 1, O: 16$)