Question

sulphuric acid is labelled 86% by weight. Density is 1.8 g/cm^3. what is the molarity?

Answer 1

Answer:

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Explanation:

Thus 8.78 moles of H2SO4 are present in 555.55 mL or 0.55555 liter. Molarity = 8.78/0.55555 = 15.80 M...........

Answer 2

w/w % of Sulphuric acid = 86% .

This means that each 100 g of solution contains 86 g of Sulphuric acid.

Hence , mass of Sulphuric acid = 86 g .

Mass of solution = 100 g (when mass of solution is not given , we take it as 100g)

Mass of water = Mass of solution - Mass of Sulphuric acid

= 100 - 86

= 14 g

Number of moles of Sulphuric acid = $\rm \frac{Given\ mass\ of\ Sulphuric\ acid}{Molar\ mass\ of\ Sulphuric\ acid}$

Molar mass of Sulphuric acid $\rm H_2SO_4$ = 2× Mass of H + Mass of sulphur + 4 × Mass of oxygen

= 2 × 1 + 32 + 4 × 16

= 2 +32 + 64

= 98 g

Number of moles of Sulphuric acid = $\rm \frac{86}{98}$

= 0.87 (approx)

Given that density of the solution is 1.8 g/cm³ .

By the formula of mass density ,

$\boxed{ \rm \: \blue {density} \: = \frac{ \orange{mass}}{\pink{volume} }}$

$\implies \rm \: 1.8 = \frac{100}{ {volume}}$

$\rm \implies \: volume = \frac{100}{1.8} ml$

$\implies \rm \: volume \: = 55.5 \: ml \:$

$\implies \rm \: volume = \frac{55.5}{1000} \\ \\ \implies \rm \: volume \: = 0.0555 \: l$

$\boxed{ \small \: \rm \blue{molarity} = \frac{ \red{number \: of \: moles \: of \: solute}}{ \purple{volume \: of \: solution \: in \: litres} }}$

$\implies\rm \: molarity = \frac{0.87}{0.0555}$

$\boxed{\huge{\therefore \rm Molarity = 15.67 M}}$