Asked by lucy
If 25cm^3 of NaOH sol containing 4.0g per litre of sol neutralised 50cm^3 of a monobasic acid,HX containing 1.8g of acid,calculate the RFM mass of acid.plz explain in detail
Answers
Answered by
DrBob222
NaOH + HX ==> NaX + H2O
M NaOH = mols/L = grams/RFM/L = 4.0/40/1 = 0.1 M
millimols NaOH = mL x M = 25 x 0.1 = 2.5
Looking at the equation, 1 mol NaOH neutralizes exactly 1 mol HX; therefore, 2.5 millimols NaOH will neutralize exactly 2.5 millimols HX or 0.025 mol HX.. Then mols = g/RFM. You know g and mols, solve for RFM
Post your work if you get stuck.
M NaOH = mols/L = grams/RFM/L = 4.0/40/1 = 0.1 M
millimols NaOH = mL x M = 25 x 0.1 = 2.5
Looking at the equation, 1 mol NaOH neutralizes exactly 1 mol HX; therefore, 2.5 millimols NaOH will neutralize exactly 2.5 millimols HX or 0.025 mol HX.. Then mols = g/RFM. You know g and mols, solve for RFM
Post your work if you get stuck.
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