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Grade: 9

                        

What weight of CaCO3 shall be neutralised by 100 ml of 0.1 M(molarity) HCl?

2 years ago

Answers : (2)

Mahima Kanawat
1010 Points
							For neutralisation , No. Of equivalent of CaCO3 = no. Of equivalent of HClNo. Of equivalent of HCL =  normality × volume                     =molarity × valency factor × volume                      = 0.1 × 1 × 0.1 = 0.01 Now equivalent of calcium carbonate should be equal to equivalent of HCL0.01= wt /equivalent wt     ( equivalent wt = molecular wt / valency factor )                            = 100 /2 = 50                                                                      0.01 = wt / 50Wt = 0.5 gm
						
2 years ago
Ankit
38 Points
							No of equvelence of HCl = N×M×V = 0.01No.  of equvelence of CaCO3=                             (wt taken) /equvelence wtequvelence wt of CaCO3=      molecular wt/
						
2 years ago
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