Harvard UniversityCHEM 5.60

20 grams of sodium nitrate is added to 250 mL of water. What is the concentration of the solution in molarity?

How would you dilute a 10M solution so you had 400 mL of a 2M solution? *

Rubbing alcohol is a 5% solution. How much alcohol would be in a full 250 mL bottle?

How many grams of salt are there in 300 mL of a 0.6M solution of salt water? 

Answer:

a) 0.941M

b) 80mL 

c) 12.5mL

d) 10.5g

Step-by-step explanation

a) 20 grams of sodium nitrate is added to 250 mL of water.

We need to find the moles NaNO₃ that are being dissolved in water

Molar mass of NaNO₃ = 84.9947 g/mol

moles of NaNO₃ = 20g/84.9947 g/mol = 0.23531moles 

We then need to find the concentration using the formula

Concentration = moles/Volume(L)

1000mL = 1L

Volume = 250 mL = 250 mL x 1L/1000mL = 0.25L

Concentration = 0.23531moles/0.25L =0.941Mol/L = 0.941M

b) 10M solution so you had 400 mL of a 2M solution

We need to find the Volume of water needed to dilute the 10M using M1V1 = M2V2

M1 = 10M and V1 = ?

M2 = 2M and V2 =400 mL

10M x V1 = 2M x 400mL

V1 = 2M x 400mL/10M = 80mL 

c) Rubbing alcohol is a 5% solution. Volume of water is 250 mL. 

This is volume by volume percent concentration, v/v%,

v/v% = Volume of Alcohol/Volume of water x 100

5% = Volume of Alcohol/250 mL x 100

Volume of Alcohol = 5 x 250/100 = 12.5mL

d) 300 mL of a 0.6M solution of salt water

We first find the moles of salt water solution 

Moles = concentration x Volume(L)

Moles of salt water = 0.6M x 300 mL x1L/1000mL= 0.18moles 

molar mass of salt water = 58.44 g/mole

Mass of salt water = moles x molar mass = 58.44 g/mole x 0.18moles = 10.5g