How many grams of $C a W O_{4}$ would contain the same mass of tungsten that is present in 569g of $F e W O_{4}$ $? (W=184)

298

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540

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487

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474

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Solution

The correct option is B $540$ g
Let the mass of $C a W O_{4}$ be w g. As given,
Mass of W in w g of $C a W O_{4}$ = mass of W in 569g of $F e W O_{4}$

Moles of W in $C a W O_{4} \times$ at. mass of W = Moles of W in $F e W O_{4} \times$ at. mass of W

As both $C a W O_{4}$ and $F e W O_{4}$ contains 1 atom of W each,
∴ moles of $C a W O_{4} \times$ at. mass of W = Moles of $F e W O_{4} \times$ at. mass of W
= $\frac{w}{288} \times 184$ = $(569 / 304) \times 184$
= $W = 539.05 g .$

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How many grams of CaWO4 would contain the same mass of tungsten that is present in 569g of FeWO4 $? (W=184)

How many grams of $C a W O_{4}$ would contain the same mass of tungsten that is present in 569g of $F e W O_{4}$ $? (W=184)