What mass of K2CrO4 is required to prepare 250cm3 of 0.020 mol dm3 solution?
[K2CrO4 = 194.2g mol-1]
-
0.97g
- 9.70g
- 19.42g
- 97.10g
The correct answer is: A
Explanation
1 mole m-3 K2CrO4 contains 194.2g mol-1 in 1000cm31 mole m-3 K2CrO4 contains 92.2g mol-1 in 500cm3
1 mole m-3 K2CrO4 contains 48.55g mol-1 in 250cm3
If n molm-3 K2CrO4 contain 48.55
0.02 molm-3 will contain (0.02/1) * (48.55/1) = 0.97g