(A) The formula for molarity is: $M = \frac{w \times 1000}{m \times V(mL)}$Where:$w = \text{mass of solute in } g$$m = \text{molar mass of solute} = 8

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(A) The formula for molarity is: $M = \frac{w \times 1000}{m \times V(mL)}$Where:$w = \text{mass of solute in } g$$m = \text{molar mass of solute} = 82.0245 \ g \ mol^{-1}$$V = \text{volume of solution} = 500 \ mL$$M = \text{molarity} = 0.375 \ M$Rearranging for $w$:$w = \frac{M \times m \times V(mL)}{1000}$$w = \frac{0.375 \times 82.0245 \times 500}{1000}$$w = 15.3796 \approx 15.38 \ g$

Class 12 Chemistry Solutions Method of expressing concentration of solution

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Calculate the mass of sodium acetate $(CH_3COONa)$ required to make $500 \ mL$ of a $0.375 \ M$ aqueous solution. The molar mass of sodium acetate is $82.0245 \ g \ mol^{-1}$. (in $g$)

A
$15.38$
B
$30.76$
C
$7.69$
D
$41.01$

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Method of expressing concentration of solution

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Explain parts per million (ppm) in brief.

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With an increase in temperature,which of the following concentration terms changes?

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Density of a $2.05 \ M$ solution of acetic acid in water is $1.02 \ g/mL$. The molality of the solution is ............. $mol \ kg^{-1}$.

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Calculate the mass of sodium acetate $(CH_3COONa)$ required to make $500 \ mL$ of a $0.375 \ M$ aqueous solution. The molar mass of sodium acetate is $82.0245 \ g \ mol^{-1}$. (in $g$)

Similar Questions

How many $cc$ of water should be added to $200 \ cc$ of $seminormal$ $NaOH$ solution to make it $decinormal$?

When $120 \ g$ of urea (molar mass $= 60 \ g/mol$) is dissolved in $1000 \ g$ of water,the density of the resulting solution is $1.15 \ g/mL$. The molarity of this solution is ........ $M$.

Explain parts per million (ppm) in brief.

With an increase in temperature,which of the following concentration terms changes?

Density of a $2.05 \ M$ solution of acetic acid in water is $1.02 \ g/mL$. The molality of the solution is ............. $mol \ kg^{-1}$.

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