given 95 weight% of HCL and Given density is 1.08g/mol
find the molarity of HCL
•If 1ml contains 1.08g HCL,
then 1000ml (1L) contains (1.08g x1000) = 1080g HCL (assuming the acid is 100% pure)
•If 100% HCL solution contains 1180g, therefore
95% HCL solution contains"= \\frac{(95 \u00d71080)}{(100)} = 1026g\/L"
But 36.5g/L of HCL = 1M,
therefore 1026g/L "= \\frac{(1026g\/L\u00d71M)}{(36.5g\/L)} = 28.11M"
Comments
Leave a comment