JEE Main202131 Aug 2021Morning ShiftChemistrySolutionsActual
The molarity of the solution prepared by dissolving 6 . 3 g of oxalic acid H 2 C 2 O 4 · 2 H 2 O in 250 mL of water in mol L - 1 is x × 10 - 2 . The value of x is _________ . (Nearest integer) [Atomic mass : H : 1 . 0 , C : 12 . 0 , 0 : 16 . 0 J ]
Correct answer
0
Step-by-step solution
M = W solute × 1000 GMM solute × V ml = 6 . 3 × 1000 126 × 250 = 0 . 2   M = 20 × 10 - 2 So X = 20