Answer to Question #97751 in General Chemistry for Jamie

Question #97751
1.00 mL of a 3.80 × 10–4 M solution of oleic acid is diluted with 9.00 mL of petroleum ether, forming solution A. Then 2.00 mL of solution A is diluted with 8.00 mL of petroleum ether, forming solution B. What is the concentration of solution B?
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Expert's answer
2019-11-01T05:44:54-0400
M1V1=M2V2M2=M1V1V2M_1 V_1 = M_2 V_2 \\ M_2 =\frac{ M_1 V_1} {V_2}

After first dilution,


M2=(3.95×104mol L1)(0.001L).0100L=3.95×105MM_2=\frac {(3.95\times10^{−4 }mol\ L^{-1} )(0.001L)} {.0100L} = 3.95\times10^{−5} M

After second dilution,

M3=(3.95×105)(0.002)0.010=7.90×106 MM_3=\frac{(3.95\times10^{-5})(0.002)}{0.010}=7.90\times10^{-6}\ M


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Comments

Assignment Expert
15.04.21, 13:01

Dear maryam albalushi post a new task

maryam albalushi
14.04.21, 21:53

1.00 mL of a 3.50 × 10–4 M solution of oleic acid is diluted with 9.00 mL of petroleum ether, forming solution A. 2.00 mL of solution A is diluted with 8.00 mL of petroleum ether, forming solution B. How many grams of oleic acid is 5.00 mL of solution B? (molar mass for oleic acid = 282 g/mol)

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