calculate the mass of Na2CO3 which have the same no. of molecules as contained in 12.3 g of MgSO4• 7 H2O [ At. masses: Na=23u , C=12u, O=16 , Mg = 24u , S=32u]
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hello,
we know that
no. of moles(n)= given mass(w)/molecular mass(m)
n=w/106 (mass of Na₂SO₃=106u)
now for MgSO₄·7H₂O
n=12.3/246 (mass of MgSO₄·7H₂O is 246u)
n=1/20
but it is given that they both have same molecules i.e
w/106=1/20
w=106/20
w=53/10
w=5.3
therefore,the mass of sodium carbonate should be 5.3 g
hope this helps,if u like it please mark it as brainliest
we know that
no. of moles(n)= given mass(w)/molecular mass(m)
n=w/106 (mass of Na₂SO₃=106u)
now for MgSO₄·7H₂O
n=12.3/246 (mass of MgSO₄·7H₂O is 246u)
n=1/20
but it is given that they both have same molecules i.e
w/106=1/20
w=106/20
w=53/10
w=5.3
therefore,the mass of sodium carbonate should be 5.3 g
hope this helps,if u like it please mark it as brainliest
ritvikjain2090ow0ydl:
please mark mine brainliest
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