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HACTEHA [7]
3 years ago
5

A 9.35g sample of a solid was placed in a 15.00mL flask. The remaining volume was filled with Benzene in which the solid is inso

luble. The solid and Benzene together weigh 17.33 g. The density of Benzene is 0.876 g/mL. What is the density of the solid?
Chemistry
2 answers:
klio [65]3 years ago
4 0

Answer:  1.58 g/ml

Explanation:

Given :Total mass= 17.33 g

Mas of solid = 9.35 g

Thus Mass of benzene = Total mass - Mass of solid= (17.33-9.35)g= 7.98 g  

Density of benzene = 0.876 g/ml

Therefore  Volume of benzene =\frac{\text {mass of benzene}}{\text {density of benzene}}=\frac{7.98g}{0.876g/ml}=
9.1 ml

Total volume = 15 ml

Volume of solid = Total volume - Volume of benzene= (15-9.1)=5.9ml

Thus Density of solid= \frac{\text {mass of solid}}{\text {volume of solid}}=\frac{9.35g}{5.9ml}=
1.58g/ml

NikAS [45]3 years ago
3 0
The problem ask to calculate the density of the solid if it has a mass of 9.35g and is place in a 15ml flask. Base on my calculation and further formulation about the density of the two, by subtracting the density of the Benzene, the solid has a density of 1.59g/ml. I hope this would help 
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14. 60. g of NaOH is dissolved in enough distilled water to make 300 mL of a stock solution. What volumes of this solution and d
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The question is incomplete, the complete question is attached below.

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Now we have to determine the volume of stock solution and distilled water mixed.

Formula used :

M_1V_1=M_2V_2

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From data (A) :

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Putting values in above equation, we get:

5M\times 20mL=1M\times V_2\\\\V_2=100mL

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From data (B) :

M_1=5M\\V_1=20mL\\M_2=1M\\V_2=?

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5M\times 20mL=1M\times V_2\\\\V_2=100mL

Volume of stock solution = 20 mL

Volume of distilled water = 100 mL - 20 mL = 80 mL

From data (C) :

M_1=5M\\V_1=60mL\\M_2=1M\\V_2=?

Putting values in above equation, we get:

5M\times 60mL=1M\times V_2\\\\V_2=300mL

Volume of stock solution = 60 mL

Volume of distilled water = 300 mL - 60 mL = 240 mL

From data (D) :

M_1=5M\\V_1=20mL\\M_2=1M\\V_2=?

Putting values in above equation, we get:

5M\times 60mL=1M\times V_2\\\\V_2=300mL

Volume of stock solution = 60 mL

Volume of distilled water = 300 mL - 60 mL = 240 mL

From this we conclude that, when 20 mL stock solution and 80 mL distilled water mixed then it will result in a solution that is approximately 1 M NaOH.

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