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# unit 2 water of crytalization

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Why bother with a post grad course - waste of time? 17-10-2016
1. i did the first two steps here and got 0.13611 mole but then didnt know how to continue. the markscheme used a mole ratio of 1:6.43 but im confused.. can someone pls explainnnn
heres a copy of the ms.. so sorry if it isnt clear

If x = 6.41 (from Mr = 120/120.1) 6.42 (from Mr = 120.3) 6.43 (from Mr = 120.4)and there is some evidence of working,award all 3 marksIf the masses of water and MgSO4 are transposed,then x = 6.96 and scores 2Answer must be to 3SFAnswer alone scores (1)n(MgSO4) = 2.55 ÷ 120.4 = 0.021179 (mol) (1)(m(H2O) = 5.00 – 2.55 = 2.45)n(H2O) = 2.45 ÷ 18 = 0.136111 (mol) (1)(Ratio 1:6.43) x = 6.43TE on calculated values above
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3. (Original post by pondsteps)
i did the first two steps here and got 0.13611 mole but then didnt know how to continue. the markscheme used a mole ratio of 1:6.43 but im confused.. can someone pls explainnnn
heres a copy of the ms.. so sorry if it isnt clear

If x = 6.41 (from Mr = 120/120.1) 6.42 (from Mr = 120.3) 6.43 (from Mr = 120.4)and there is some evidence of working,award all 3 marksIf the masses of water and MgSO4 are transposed,then x = 6.96 and scores 2Answer must be to 3SFAnswer alone scores (1)n(MgSO4) = 2.55 ÷ 120.4 = 0.021179 (mol) (1)(m(H2O) = 5.00 – 2.55 = 2.45)n(H2O) = 2.45 ÷ 18 = 0.136111 (mol) (1)(Ratio 1:6.43) x = 6.43TE on calculated values above
5.00 grams decreased to 2.55g

In other words 2.45g of water was lost.

The mass of MgSO4 remaining = 2.55g
Hence moles of MgSO4 remaining = 2.55/Mr = 2.55/120 = 0.02125

Moles of water released = 2.45/Mr = 2.45/18 = 0.1361

Ratio of water to MgSO4 = 0.1361/0.02125 : 1

Ratio of water to MgSO4 = 6.41 : 1 (3 sig figs)

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