1) If 20.0 g of MgSO47H2O is thoroughly heated, what mass of anhydrous magnesium

Question

1) If 20.0 g of MgSO47H2O is thoroughly heated, what mass of anhydrous magnesium sulfate will remain? 2) How many molecules are contained in a 4.50-g sample of dimethylmercury? What is the percentage of mercury (by mass) in the sample? 3) What is the molar mass of pentane if 4.18 × 1016 molecules of pentane weigh 5.00 g? 4) What is the molar mass of pentane if 4.18 × 1016 molecules of pentane weigh 5.00 g? 139 g/mol 288 g/mol 347 g/mol 72.0 g/mol 5) Determine the mass of oxygen in a 4.8 g sample of Al(NO3)3.

Explanation / Answer

1) The molar mass of MgSO4. 7H2O is 246.52 g/mol. Out of this mass of water is 7 x 18.015 g = 126.11 g

therefore mass of anhydrous MgSO4 in 246.52 g of MgSO4.7H2O = 246.52 g - 126.11 g = 120.41 g

therefore 20 g of MgSO4.7H2O will give (120.41/ 246.52) x 20 g of anhydrous MgSO4 = 9.77 g of anhydrous MgSO4

2) Molecular Weight of Dimethyl Mercury (CH3HgCH3) = 230.66 g

This means 6.023 x 1023 molecules (avagadro no.) has a mass of 230.66 g

So 4.50 g of sample has 4.50 x 6.023 x 1023 / 230.66molecules = 11.75 x 1021 molecules

Now Atomic wt of Hg is 200.59 g

Hence % of Hg by mass = (200.59/ 230.66) x100 = 86.96 %

3) 4.18 × 1016 molecules of pentane weigh 5.00 g

therefore 6.023 x 1023 molecules (avagadro no.) of pentane has a mass of (5.00 x 6.023 x 1023 )/ 4.18 × 1016 g

= 72.0 x 106 g = 72.0 g

there molar mass of pentane is 72.0 g/ mole.

4) same question as no 3.

5) Al(NO3)3 has a Molar mass: 212.996 g/mol. Out of this mass of oxygen = 9 x 16 = 144 g

that is 212.996 g of Al(NO3)3 has 144.0 g of oxygen

so 4.8 g of Al(NO3)3 will have 144.0 x 4.8/ 212.996 = 3.245 g of oxygen.

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