According to the source below, the solubility of sulfanilamide in 95% ethyl alcohol at 78°C is 210 mg/mL. Since 0.1 g = 100 mg, we can set up a proportion:
(210 mg) / (1 mL) = (100 mg) / (x mL) Solving, x = 0.48 mL of 95% ethyl alcohol will be required.
I do not know previously the solubility of sulfanilamide in 95% ethyl alcohol. Let us accept the solubility you quoted here.
100/210 = 0.47619047619.. ≈ 0.48 (ml)
at 0C, the amount of sulfanilamide remains in the solution is: 14*(100/210) = 6.67 (mg), since you only have 0.48 ml solution.
The volume of the solution will change a little by cooling from 78C to 0C. You may also consider this volume change if you have data.
For Ethanol:
D = m / V
D = 3.9 g / 5 mL
D = 0.78 g/mL
For benzene:
D = 4.4 g / 5 mL
D = 0.88 g/mL
benzene has a <span>higher density.
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hope this helps!
A) LiAlH4
b) Ph(3)SnH
c)Na in dry ether(Wurtz reaction)
Answer: -
15.55 M
35.325 molal
Explanation: -
Let the volume of the solution be 1000 mL.
Density of nitric acid = 1.42 g/ mL
Total Mass of nitric acid Solution = Volume of nitric acid x Density of nitric acid
= 1000 mL x 1.42 g/ mL
= 1420 g.
Percentage of HNO₃ = 69%
Amount of HNO₃ = 
= 979.8 g
Molar mass of HNO₃ = 1 x 1 + 14 x 1 + 16 x 3 = 63 g /mol
Number of moles of HNO₃ = 
= 15.55 mol
Molarity is defined as number of moles per 1000 mL
We had taken 1000 mL as volume and found it to contain 15.55 moles.
Molarity of HNO₃ = 15.55 M
Mass of water = Total mass of nitric acid solution - mass of nitric acid
= 1420 - 979.8
= 440.2 g
So we see that 440.2 g of water contains 15.55 moles of HNO₃
Molality is defined as number of moles of HNO₃ present per 1000 g of water.
Molality of HNO₃ = 
= 35.325 molal