Answer:
MgSO4.7H2O
Explanation:
Let the formula for the hydrated magnesium sulphate be MgSO4.xH2O
Mass of the hydrated salt (MgSO4.xH2O) = 12.845g
Mass of anhydrous salt (MgSO4) = 6.273g
Mass of water molecule(xH2O) = Mass of the hydrated salt — Mass of anhydrous salt = 12.845 — 6.273 = 6.572g
Now,we can obtain the number of mole of water molecule present in the hydrated salt as follows:
Molar Mass of hydrated salt (MgSO4.xH2O) = 24 + 32 + (16x4) + x(2 + 16) = 24 + 32 + 64 + x(18) = 120 + 18x
Mass of xH2O/ Molar Mass of MgSO4.xH2O = Mass of water / mass of hydrated salt
18x/120 + 18x = 6.572/12.845
Cross multiply to express in linear form
18x x 12.845 = 6.572(120 + 18x)
231.21x = 788.64 + 118.296x
Collect like terms
231.21x — 118.296x = 788.64
112.914x = 788.64
Divide both side by 112.914
x = 788.64 /112.914
x = 7
Therefore the formula for the hydrated salt (MgSO4.xH2O) is MgSO4.7H2O
Ch = sh = why and find the stuff
Answer:
[H₂SO₄] = 6.07 M
Explanation:
Analyse the data given
8.01 m → 8.01 moles of solute in 1kg of solvent.
1.354 g/mL → Solution density
We convert the moles of solute to mass → 8.01 mol . 98g /1mol = 785.4 g
Mass of solvent = 1kg = 1000 g
Mass of solution = 1000g + 785.4 g = 1785.4 g
We apply density to determine the volume of solution
Density = Mass / volume → Volume = mass / density
1785.4 g / 1.354 g/mL = 1318.6 mL
We need this volume in L, in order to reach molarity:
1318.6 mL . 1L / 1000mL = 1.3186 L ≅ 1.32L
Molarity (mol/L) → 8.01 mol / 1.32L = 6.07M