(B) The number of moles of solute is calculated using the formula: $n = M \times V$ (where $M$ is molarity and $V$ is volume in liters).For $1000 \ mL

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(B) The number of moles of solute is calculated using the formula: $n = M \times V$ (where $M$ is molarity and $V$ is volume in liters).For $1000 \ mL$ of $1 \ M \ NaOH$,$n = 1 \ M \times 1 \ L = 1 \ mol$.Checking the options:$(1)$ $0.2 \ M \times 0.2 \ L = 0.04 \ mol$.$(2)$ $1 \ M \times 0.2 \ L = 0.2 \ mol$ and $0.2 \ M \times 1 \ L = 0.2 \ mol$. Since $0.2 = 0.2$,this statement is true.$(3)$ $0.2 \ M \times 0.1 \ L = 0.02 \ mol$.$(4)$ $0.2 \ M \times 2 \ L = 0.4 \ mol$.

Class 12 Chemistry Solutions Method of expressing concentration of solution

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Which of the following statements is true?
$(1)$ $200 \ mL$ of $0.2 \ M \ NaOH$ has the same number of moles of solute as $1000 \ mL$ of $1 \ M \ NaOH$.
$(2)$ $200 \ mL$ of $1 \ M \ NaOH$ has the same number of moles as $1000 \ mL$ of $0.2 \ M \ NaOH$.
$(3)$ $100 \ mL$ of $0.2 \ M \ NaOH$ has the same number of moles as $1000 \ mL$ of $1 \ M \ NaOH$.
$(4)$ $2000 \ mL$ of $0.2 \ M \ NaOH$ has the same number of moles as $1000 \ mL$ of $1 \ M \ NaOH$.

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A
$1$
B
$2$
C
$3$
D
$4$

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$10.6 \, g$ of a substance of molecular weight $106$ was dissolved in $100 \, mL$. $10 \, mL$ of this solution was pipetted out into a $1000 \, mL$ flask and made up to the mark with distilled water. The molarity of the resulting solution is

$A$ 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 $X$?

If the concentration of glucose $(C_{6}H_{12}O_{6})$ in blood is $0.72 \ g \ L^{-1}$,the molarity of glucose in blood is $..... \times 10^{-3} \ M$. (Nearest integer)(Given: Atomic mass of $C=12, H=1, O=16 \ u$)

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If the concentration of glucose $(C_{6}H_{12}O_{6})$ in blood is $0.72 \ g \ L^{-1}$,the molarity of glucose in blood is $..... \times 10^{-3} \ M$. (Nearest integer)
(Given: Atomic mass of $C=12, H=1, O=16 \ u$)