chemistry

posted by .

calculate the mass of Cuso4.5H2O required to prepare 500 cm3 of a 0.400 m/dm-3 solution

  • chemistry -

    How many mols do you want? That's M x L = 0.400 x 0.500 = ?
    Then mols = g/molar mass. You know molar mass and mols, solve for mols.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. chemistry

    Calculate the mass of the solid you must measure out to prepare 100 mL of 0.025 M CuSO4. Note that this salt is a hydrate, so its formula is CuSO4⋅5H2O. You must include the waters of hydration when calculating the formula weights. …
  2. Chemistry

    how much CuSO4.5H2O is required to prepare 100 ml of a .050M aqueous stock solution?
  3. CHEMISTRY

    Find the amount of CuSO4.5H2O needed to prepare 500 mL of 0.25 M solution.
  4. chemistry

    i find it confusing on how to do this question.. what mass of solute is needed to prepare 400 mL of 0.850 M CuSO4 from CuSO4.5H2O
  5. Analytical chemistry

    How many grams of solid CuSO4. 5H2O are needed to prepare a 500 ml of a 1000 ppm solution of Cu?
  6. science

    calculate the mass of Cuso4.5H2O required to prepare 500 cm3 of a 0.400 m/dm-3 solution
  7. chemistry

    How many grams of CuSO4 . 5H2O are needed to prepare 100ml of a 0.10M solution?
  8. Chemistry

    A student was asked to prepare a sample of copper(ii)sulphate crystals(CUSO4.5H2O) given 5g of of copper (ii)oxide (CUO) as the starting material.if 10g of dry crystals of CUSO4.5H2O were obtained,calculate the percentage yield?
  9. Chemistry

    A chemist dissolves 0.096 g of CuSO4 · 5 H2O in water and dilutes the solution to the mark in a 500-mL volumetric flask. A 10.8-mL sample of this solution is then transferred to a second 500-mL volumetric flask and diluted. What is …
  10. Chemistry

    A chemist dissolves 0.096 g of CuSO4 · 5 H2O in water and dilutes the solution to the mark in a 500-mL volumetric flask. A 10.8-mL sample of this solution is then transferred to a second 500-mL volumetric flask and diluted. What is …

More Similar Questions