### Molarity calculation - Random types

4. Moles and Molarity 4.2) Molarity 4.2.1) Molarity - general

4.2.1.1) Molarity - general Normal 1

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109.88 g of sodium phosphate dibasic heptahydrate is dissolved in 200 mL of solution. If the molecular weight of sodium phosphate dibasic heptahydrate is 268, then determine the molarity (M) of the solution Click on the button below to see the answer and explanations

lb equals 2.05 M kg

Molarity is the number of moles (n) dissolved per liter of solution (V). Again, mole is the weight in gram (g) divided by the molecular weight (a).
In this example, the volume of the solution is 200 mL, or 0.2 L.

 Using formula method:
M=g/(aV)=(109.88\quad g)/(268\quad (g/(mol)) × 0.2\quad L)
therefore M = 2.05 (mol)/L=2.05 without unit, Ans.

 Using conceptual method:
First, count the number of moles (n) of sodium phosphate dibasic heptahydrate present in this solution:
=(weight\quad i n \quad g)/(mol ecu lar\quad wt)=(109.88\quad g)/(268 \quad (g/(mol)))=0.41 \quad mol
Then, calculate how many moles will be present in 1L of solution. Using equation, we can write:
(0.41\quad mol e)/(200 \quad mL)=x/(1000 \quad mL) therefore x = 2.05 Ans.