What volume of 12.0 M HCI must be used to prepare 250 mL of 5.50 M HCI? What is

Question

What volume of 12.0 M HCI must be used to prepare 250 mL of 5.50 M HCI? What is the Molarity of a sample of NaOH if a titration requires 35.0 mL of 4.00 M HCI to neutralize 25.0 mL of the NaOH solution? What will be the new molarity if 195 mL of 5.0 M HCI is diluted to a volume of 225mL? What is the molarity of a solution if 25.0 g of NaCI is dissolved in enough water to give a volume of 325 mL? Calculate the mass % of CaCl_2 in a solution that has 6.00 g of CaCL_2 in 575 g of H_2 O. In a can of soda, the water is the____(solvent or solute). the 9 tablespoons of sugar are the____(solvent or solute). and the soda itself is the___(solute, solvent, or solution). Predict if each of the following would dissolve in gasoline (C_8 H_18). (a) H_2 O (b) oil (C_14 H_30) (c) NaCl Explain your reasoning above, on if the following would dissolve in gasoline (C_8 H_18). (a) H_2 O (b) oil (C_14 H_30) (c) NaCl

Explanation / Answer

1)

concentration of HCl used = 12 M

concentration of HCl prepared = 5.50 M

volume = 250 mL

M1 V1 = M2 V2

12 x V = 5.50 x 250

V = 114.6 mL

volume of HCl used = 114.6 mL

2)

35 x 4 = 25 x M

Molarity of NaOH = 5.6 M

3)

195 x 5 = 225 x M

new Molarity = 4.33 M

4)

moles of NaCl = 25 / 58.5 = 0.427

volume = 325 mL = 0.325 L

Molarity = 0.427 / 0.325

             = 1.31 M

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