Question

10.0 mL of the aqeuous ammonium chloride solution from question 1 above is diluted by adding 15.0 mL of distilled water. What is the new Molarity of this diluted solution? Give your answer to the correct number of significant figures. Type in only the numbers. Do not include the molarity unit (M or moles/L).

Note: the answer will be a molarity less than 1. Make sure to type in the leading zero. For example, if your answer is 0.111, type in 0.111 with the "zero point", and NOT .111

Question text

This is question one

11.3 grams of ammonium chloride (NH₄Cl) are dissolved in enough water to make 155 mL of solution. What is the resulting Molarity of the aqeuous solution? Give your answer to the correct number of significant figures. Type in only the numbers. Do not include the molarity unit (M or moles/L). Hint: you need to solve for the molar mass of ammonium chloride to do this calculation.

Answer: 1.36

Answer 1

Molar mass of NH4Cl = 53.5 g/mol

Number of moles = Mass/Molar mass = 11.3/53.5 = 0.2112 moles

Molarity = Number of moles/Volume of solution (in L) = 0.2112/0.155 = 1.36 M (Answer to Question 1)

Next Question:

Number of moles will remain the same before and ater the dilution

M1 = 1.36 M

V1 = 10.0 ml

V2 = 10.0 + 15.0 (water added volume) = 25.0 ml

M1V1 = M2V2

1.36 * 10.0/1000 = M2 * 25.0/1000

M2 = 1.36/2.5 = 0.544 M

Hence the final answer will be 0.544 M

Note - Post any doubts/queries in comments section.