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