Answer to Question #138005 in General Chemistry for Juliana

Question

Resource 21: Dilutions (M1xV1 = M2xV2)

(Ref WS Q1) Calculate the volume (in L) of 5.0 M lithium sulfate solution needed to make 2.0 L of 1.0 M solution
(Ref WS Q3) Calculate the volume in mL required to prepare 500. mL of a 1.77 M H2SO4 solution form an 18.0 M H2SO4

(Ref WS Q4) Calculate the volume in mL of a 0.05 M HCl solution made by diluting 250 mL of 10 M HCl

(Ref WS Q6) Calculate the molarity when 250 mL of a 0.10 M solution of Na2CO3 is diluted with 750 mL of water

(Ref WS Q8) Calculate the molarity of 65.5 mL HCl stock solution is used to make 450. mL of a 0.675 M HCl dilution

(Ref WS Q9) Calculate the molarity when 95 mL evaporates from 345 mL of a 1.5 M NaCl solution

Answer 1

QUESTION # 138005

Resource 21: Dilutions (M1xV1 = M2xV2)

(Ref WS Q1) Calculate the volume (in L) of 5.0 M lithium sulfate solution needed to make 2.0 L of 1.0 M solution

ANSWER

M1xV1 = M2xV2

5.0M× V1= 1.0Mx2.0L

V1=(1.0Mx2.0L)÷ 5.0M

           =0.4L

(Ref WS Q3) Calculate the volume in mL required to prepare 500. mL of a 1.77 M H2SO4 solution form an 18.0 M H2SO4

ANSWER

M1xV1 = M2xV2

18.0M× V1= 1.77Mx500mL

V1=(1.77Mx500mL)÷ 18.0M

           =49.17mL

(Ref WS Q4) Calculate the volume in mL of a 0.05 M HCl solution made by diluting 250 mL of 10 M HCl

ANSWER

M1xV1 = M2xV2

10M×250ml=0.05M× V2

V2=(10M×250ml)÷ 0.05M

           =50000mL

(Ref WS Q6) Calculate the molarity when 250 mL of a 0.10 M solution of Na2CO3 is diluted with 750 mL of water

ANSWER

M1xV1 = M2xV2

V2=750+250=1000Ml

0.10M×250ml=M2×1000ml

M2=(0.10M×250ml)÷ 1000ml

           =0.025M

(Ref WS Q8) Calculate the molarity of 65.5 mL HCl stock solution is used to make 450. mL of a 0.675 M HCl dilution

ANSWER

M1xV1 = M2xV2

M1×65.5mL=0.675M×450mL

M1=(0.675M×450mL)÷ 65.5mL

           =4.60M

(Ref WS Q9) Calculate the molarity when 95 mL evaporates from 345 mL of a 1.5 M NaCl solution

ANSWER

M1xV1 = M2xV2

V2=345-95=250mL

1.5M×345mL=M2×250mL

M2=(1.5M×345mL)÷ 250Ml

           =2.07M