The ratio of moles of Na2CO3 to total number of moles in a 100 mL aqueous solution of Na2CO3 is 1:5. If 90 g of water is present in 100 mL of solution, find the molarity of the solution.
A
12.5 M
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
1.2580 M
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
0.0125 M
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
0.1538 M
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A 12.5 M Let the moles of sodium carbonate be denoted as ‘a’ mol According to the given data: 90 g of water is present in 100 mL of the solution. The given ratio of moles of sodium carbonate to the total number of moles in solution does not change. Number of moles of water present in 100 mL of the solution =9018=5mol molesofsodiumcarbonatetotalnumberofmoles=15 15=molesofsodiumcarbonatemolesofsodiumcarbonate+molesofwater ⇒15=aa+5 On solving, a = 1.25 mol molarity=molesofsolutevolumeofsolutioninL molarity=1.250×1000100=12.50M