Answer:
0.29mol/L or 0.29moldm⁻³
Explanation:
Given parameters:
Mass of MgSO₄ = 122g
Volume of solution = 3.5L
Molarity is simply the concentration of substances in a solution.
Molarity = number of moles/ Volume
>>>>To calculate the Molarity of MgSO₄ we find the number of moles using the mass of MgSO₄ given.
Number of moles = mass/ molar mass
Molar mass of MgSO₄:
Atomic masses: Mg = 24g
S = 32g
O = 16g
Molar mass of MgSO₄ = [24 + 32 + (16x4)]g/mol
= (24 + 32 + 64)g/mol
= 120g/mol
Number of moles = 122/120 = 1.02mol
>>>> From the given number of moles we can evaluate the Molarity using this equation:
Molarity = number of moles/ Volume
Molarity of MgSO₄ = 1.02mol/3.5L
= 0.29mol/L
IL = 1dm³
The Molarity of MgSO₄ = 0.29moldm⁻³
having a lesser value by a process of depletion
Answer:
C₅ H₁₂ O
Explanation:
44 g of CO₂ contains 12 g of C
30.2 g of CO₂ will contain 12 x 30.2 / 44 = 8.236 g of C .
18 g of H₂O contains 2 g of hydrogen
14.8 g of H₂0 will contain 1.644 g of H .
total compound = 12.1 out of which 8.236 g is C and 1.644 g is H , rest will be O
gram of O = 2.22
moles of C, O, H in the given compound = 8.236 / 12 , 2.22 / 16 , 1.644 / 1
= .6863 , .13875 , 1.644
ratio of their moles = 4.946 : 1 : 11.84
rounding off to digits
ratio = 5 : 1 : 12
empirical formula = C₅ H₁₂ O
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