What is the mass of solute in 500\(cm^{3}\) of 0.005\(moldm^{-3}\) \(H_{2}SO_{4}\)? ( H = 1, S = 32.0, O=16.0)
- 0.490g
- 0.049g
-
0.245g
- 0.0245g
The correct answer is: C
Explanation
Using
n = cv where n = no of mole, c = molar concentration(mol/dm3) and v = volume( dm3)
volume = 500\(cm^{3}\) = 0.5dm3, c = 0.005\(moldm^{-3}\)
number of mole(n) = c x v = 0.5 X 0.005 = 0.0025mol.
but n = \(\frac{mass}{ molar mass}\) = 0.0025 = \(\frac{mass}{ 98}\) (where 98g/mol is the molar mass of H2SO4)
mass of H2SO4 = 0.0025 x 98 = 0.245g.