Question

कैलकुलेट द नंबर ऑफ एलुमिनियम एटम्स प्रेजेंट इन 45 ग्राम ऑफ alcl3​

Answer 1

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koi baath nahi, ye lo apke answers

Molar mass of aluminum = 27

molar mass of AlCl₃ = 133.5 g

This means that 133.5 g of AlCl₃ will contain 27 g of Aluminum

no. of moles of AlCl₃ in 45 g = 45/133.5 = 0.38 moles (approx)

so, 0.38 moles will contain:   0.38 x 27 = 10.26 g of Aluminum

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