Question

The molarity of a sulphuric acid sample having density 2.12 g/ml and mass percentage 49% (w/w) is equal to

Answer 1

Explanation:

V=m/d. V=49/2.12=23.11L,. then,. Molarity=w×1000/m×V,. 51000/98×23.11=51000/2264.7=22.51m

Answer 2

Given:-

» Density of Sulphuric acid = 2.12 g/ml

» Mass percentage of Sulphuric acid = 49 %

To find:-

Molarity of sulphuric acid

Solution:-

» Molar mass of sulphuric acid = 98 mol

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Chemical formula of Sulphuric acid

$\leadsto\purple{\sf H_2SO_4}$

$\purple{\sf Molar\: mass\: of\: H_2SO_4}$

$\dashrightarrow\sf 2× Atomic \: mass\: of\: H+ Atomic \: mass\: of \: S + 4×Atomic\: mass\: of \: O$

$\dashrightarrow\sf 2×1+32+4×16$

$\dashrightarrow\sf 2+32+64$

$\dashrightarrow\sf 34+64$

$\dashrightarrow\sf 98 \: mol$

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Molarity is given

$\leadsto\large\green{\sf M=\frac{\% x× d ×10}{M.M}}$

Where,

M = molarity

% x = % w/W ( Mass Percentage)

M. M = Molecular mass of sulphuric acid

$\longrightarrow\sf \frac{49×2.12×10}{98}$

$\longrightarrow\sf \frac{490×2.12}{98}$

$\longrightarrow\sf 10.6 \: M$

Option B is correct.