Question

calculate the amount of Na2CO3(molecular weight of Na2CO3 is 106) required to prepare 50mM solution of 10ml?​

Answer 1

Answer:

You should add 5.299 grams of Sodium Bicarbonate in the 100 ml of solvent

you can calculate it using molarity relation

molarity = No. of moles / Liter of solution

Here, you have concentration , volume and molecular mass of Sodium bicarbonate

M = 0.1 , Vol= 0.100 Ltrs , lets assume number of moles = x (remember the units should be in same unit system)

substituting in above formula stated

we got x = 0.05 , now we need 0.05 molecules of sodium bicarbonate to be added in 100 ml so

now change the number of moles to grams

as we know 1 mole of Na2CO3 have molecular mass = 105.98

hence 0.05 moles will be = 0.05 x 105.98 = 5.299 grams

Answer 2

Answer:

0.053 g or 53 mg

Explanation:

1 mole = 1000 millimole

m mole = 50 millimole

m = 0.05

Molarity = weight of substance * 1000 / molecular weight * volume in mL

0.05 = x * 1000 / 106 * 10

x * 1000 = 53

x = 53 / 1000 grams

x = 0.053 g or 53 mg