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3 years ago

How much water should be added to 80 grams of common salt so as to obtain 20% salt solution

1 answer:

3 0

Concentration of solution is 20%

Mass of solute( common salt) is 80g

take the amount of water as x

mass of the solvent = (80+x)

We know, concentration of solution = (mass of solute/mass of solution)×100

20=(80/80+x)×100

80+x=80×5 ( we devide the both side by 20)

x= 400-80

x=320 g

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3 years ago

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You measure salt water in a tank to have a density of 1.02 g/mL. A balloon weighs 2.0 g and you weights have a mass of 30.0 g ea

Elden [556K]

Weight of the balloon = 2.0 g

Six weights each of mass 30.0 g is added to the balloon.

Total mass of the balloon = 2.0 g + 6*30.0 g = 182 g

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Calculating the volume from mass and density:

$182g*\frac{mL}{1.02g} =178mL$

Converting the volume from mL to cubic cm:

$178 mL * \frac{1cm^{3} }{1mL} =178cm^{3}$

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3 years ago

3mL of cyclohexanol (density = 0.9624 g/mL, Molecular weight = 100.158 g/mol) reacts with excess sulfuric acid to produce cycloh

TiliK225 [7]

Answer:

$n_{C_6H_{10}}=0.03molC_6H_{10}$

Explanation:

Hello,

In this case the undergoing chemical reaction is shown on the attached picture whereas cyclohexanol is converted into cyclohexene and water by the dehydrating effect of the sulfuric acid. Thus, for the starting 3 mL of cyclohexanol, the following stoichiometric proportional factor is applied in order to find the theoretical yield of cyclohexene in moles:

$n_{C_6H_{10}}=3mLC_6H_{12}O*\frac{0.9624gC_6H_{12}O}{1mLC_6H_{12}O}*\frac{1molC_6H_{12}O}{100.158gC_6H_{12}O}*\frac{1molC_6H_{10}}{1molC_6H_{12}O} \\n_{C_6H_{10}}=0.03molC_6H_{10}$

Besides, the mass could be computed as well by using the molar mass of cyclohexene:

$m_{C_6H_{10}}=0.03molC_6H_{10}*\frac{82.143gC_6H_{10}}{1molC_6H_{10}} \\\\m_{C_6H_{10}}=2.4gC_6H_{10}$

Even thought, the volume could be also computed by using its density:

$V_{C_6H_{10}}=2.4gC_6H_{10}*\frac{1mLC_6H_{10}}{0.811gC_6H_{10}} \\V_{C_6H_{10}}=3mLC_6H_{10}$

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3 years ago

How much water should be added to 80 grams of common salt so as to obtain 20% salt solution​

How much water should be added to 80 grams of common salt so as to obtain 20% salt solution