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)
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A.
0.490g -
B.
0.049g -
C.
0.245g -
D.
0.0245g
Correct Answer: Option 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}) = (2×1) + 32 + (16×4) = 98)
(mass in gdm^{-3} = 0.980gdm^{-3})
(1000cm^{3} = 0.980g)
(500cm^{3} = frac{0.980}{1000} times 500 = 0.490g)