2M
Explanation:
Given that:
V1 = 50mL
M1 = 40M
V2 = 1000mL
unknown:
M2 = ?
Solution:
Concentration is the number of moles solute in a solution.
Since the number of moles in the solute remains the same;
Concentration = 
number of moles = concentration x volume
Number of moles before and after the dilution remains the same;
M1 x V1 = M2 x V2
M1 is initial concentration
V1 is initial volume
M2 is final concentration
V2 is final volume
40 x 50 = M2 x 1000
M2 = 2M
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Explanation:
Mass of compounds = Moles of compound × Molecular mass of compound
a) Moles of LiCl = 2.345 mol
Molecular mass of LiCl = 42.5 g/mol
Mass of 2.345 moles of LiCl = 2.345 mol × 42.5 g/mol = 99.6625 g
b) Moles of acetylene = 0.0872 mol
Molecular mass of acetylene= 26 g/mol
Mass of 0.0872 moles acetylene= 0.0872 mol × 26 g/mol = 2.2672 g
c) Moles of sodium carbonate= 
Molecular mass of sodium carbonate= 106 g/mol
Mass of
sodium carbonate
=
= 3.498 g
d) Moles of fructose = 
Molecular mass fructose= 180 g/mol
Mass of
fructose
= 
e) Moles of 
Molecular mass of 
Mass of
fructose
= 
The reaction is
CaC₂(s) + 2H₂O (l) -----> Ca(OH)₂ (s) + C₂H₂ (g)
As we have data of gas ethyne (or acetylene), C₂H₂
We can calculate the moles of acetylene and from this we can estimate the mass of calcium carbide taken
the moles of acetylene will be calculated using ideal gas equation
PV =nRT
R = gas constant = 0.0821 Latm/molK
T = 385 K
V = volume = 550 L
P = Pressure = 1.25 atm
n = moles = ?
n = PV /RT = 1.25 X 550 / 0.0821 X 385 = 21.75 mol
As per balanced equation these moles of acetylene will be obtained from same moles of calcium carbide
moles of calcium carbide = 21.75mol
molar mass of CaC₂ = 40 + 24 = 64
mass of CaC₂ = moles X molar mass = 21.75 X 64 = 1392g