Question #26315

How many moles of hydrosulphate acid are there in 27.0 g of hydrosulphate acid if the density of hydrosulphate acid is 1.84 g/cubic centimeter. what is the concentration in %m/m of 250 cubic centimeter in weighing 350 g?

Expert's answer

To find the amount of H_{2}SO_{4} in moles we should use the formula

n(moles) = m(g)/M_{W}(g/mol)

M_{W}(H_{2}SO_{4}) = 2 x 1 + 32 + 16 x 4 = 98 g/mol

n(H_{2}SO_{4}) = 27.0/98.0 = 0.276 mol

The concentration of H_{2}SO_{4} is

w(%) = m(H_{2}SO_{4}, g)/m(solution of H_{2}SO_{4}, g) x 100%

The weight of 250 cm^3 of solution is

m(H_{2}SO_{4}) = V(cm^3) x p(g/cm^3) = 250 x 1.84 = 460 g

w(%) = 350 / 460 x 100% = 76.1 %

Answer:

n(H_{2}SO_{4}) = 0.276 mol

w(%) = 76.1 %

n(moles) = m(g)/M

M

n(H

The concentration of H

w(%) = m(H

The weight of 250 cm^3 of solution is

m(H

w(%) = 350 / 460 x 100% = 76.1 %

Answer:

n(H

w(%) = 76.1 %

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