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)
A. 0.490g
B. 0.049g
C. 0.245g
D. 0.0245g
Correct Answer: A
Explanation
\(500cm^{3} = 0.005moldm^{-3}\)
\(1000cm^{3} = \frac{0.005}{500} \times 1000cm^{3} = 0.010moldm^{-3}\)
Molar concentration = \(\frac{mass in gdm^{-3}}{molar mass in gmol^{-1}}\)
\(0.01moldm^{-3} = \frac{mass in gdm^{-3}}{98.0gmol^{-1}}\)
(Molar mass of \(H_{2}SO_{4}\) = (2x1) + 32 + (16x4) = 98)
\(mass in gdm^{-3} = 0.980gdm^{-3}\)
\(1000cm^{3} = 0.980g\)
\(500cm^{3} = \frac{0.980}{1000} \times 500 = 0.490g\)