The dilution formula can be used to find the volume needed
c1v1 = c2v2
Where c1 is concentration and v1 is volume of the concentrated solution
And c2 is concentration and v2 is volume of the diluted solution to be prepared
c1 - 0.33 M
c2 - 0.025 M
v2 - 25 mL
Substituting these values in the equation
0.33 M x v1 = 0.025 M x 25 mL
v1 = 1.89 mL
Therefore 1.89 mL of the 0.33 M solution needs to be diluted up to 25 mL to make a 0.025 M solution
Explanation:
Molar mass
The mass present in one mole of a specific species .
The molar mass of a compound , can easily be calculated as the sum of the all the individual atom multiplied by the number of total atoms .
(a) S₈
Molar mass of of the atoms are -
sulfur, S = 32 g/mol.
Molar mass of S₈ = 8 * 32 g/mol. = 256 g/mol.
(b) C₂H₁₂
Molar mass of of the atoms are -
Hydrogen , H = 1 g/mol
Carbon , C = 12 g/mol
Molar mass of C₂H₁₂ = ( 2 * 12 ) + (12 * 1 ) = 36 g /mol
(c) Sc₂(SO₄)₃
Molar mass of of the atoms are -
sulfur, S = 32 g/mol.
oxygen , O = 16 g/mol.
scandium , Sc = 45 g/mol.
Molar mass of Sc₂(SO₄)₃ = (2 * 45 ) + ( 3 *32 ) + ( 12 * 16 ) = 378 g /mol
(d) CH₃COCH₃ (acetone)
Molar mass of of the atoms are -
Carbon , C = 12 g/mol
oxygen , O = 16 g/mol.
Hydrogen , H = 1 g/mol
Molar mass of CH₃COCH₃ (acetone) = (3 * 12 ) + ( 1 * 16 ) + ( 6 * 1 ) = 58g/mol
(e) C₆H₁₂O₆ (glucose)
Molar mass of of the atoms are -
Carbon , C = 12 g/mol
oxygen , O = 16 g/mol.
Hydrogen , H = 1 g/mol
Molar mass of C₆H₁₂O₆ (glucose) = ( 6 * 12 ) + ( 12 * 1 ) + ( 6 * 16 ) = 108g/mol.
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Friction .......................................................................................
